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1 

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>:*- 


*,f 


BOWSER^S  MATHEMATICS. 

ACADEMIC  ALGEBRA.    With  numerc  is  Examples. 

COLLEGE  ALGEBRA.    With  nutnerous  Examples. 

PLANE  AND  SOLID  GEOMETRY.  With  numerous  Exer- 
ciaea. 

ELEMENTS  OF  PLANE  AND  SPHERICAL  TRIGONOME- 
TRY.    With  numerous  Examples. 

A  TREATISE  ON  PLANE  AND  SPHERICAL  TRIGONOME- 
TRY,  and  its  appliicationg  to  Astronomy  and  Geodesy. 

With  numerous  Examples. 

AN  ELEMENTARY  TREATISE  ON  ANALYTIC  GEOMETRY, 
embracing  Plane  Geometry,  and  an  Introduction  to 
Geometry  of  Three  Dimensions. 

AN  ELEMENTARY  TREATISE  ON  THE  DIFFERENTIAL 
AND  INTEGRAL  CALCULUS.  With  numi;iouB  Exam- 
ples. 

AN  ELEMENTARY  TREATISE  ON  ANALYTIC  MECHANICS. 

With  numerous  Examples. 

AN    ELEMENTARY     TREATISE     ON    HYDROMECHANICS. 

With  numerous  Examples. 


* 


A   TREATISE 


ON 


PLANE   AND   SPHERICAL 


TRiaON^OMETEY, 


AND  ITS  APPLICATIONS  TO 


ASTROE^OMY  AND   GEODESY, 


WITH 


NUMEROUS    EXAMPLES. 


BY 


EDWARD   A.   BOWSER,   LL.D., 

Pkofkssob  of  Mathematics  and  Enginbeuing  in  Rutgers  College. 


BOSTON,  U.S.A. : 
PUBLISHED  BY   D.  C.  HEATH  &   CO. 

1892. 


I 


COPTRIOHT,   1892, 

By  E.  A.  BOWSER. 


TvrOGRAPHY    BY  J.    S.    CUSHING  &   Co.,    BoSTON,    U.S.A. 

Presswork  by  Berwick  &  Smith,  Boston,  U.S.A. 


1 

i 


"-""--,- "  niTi 


PEEFACE. 


The  present  treatise  on  Plane  and  Spherical  Trigo- 
nometry is  designed  as  a  text-book  for  Colleges,  Scien- 
tific Schools,  and  Institutes  of  Technology.  The  aim 
has  been  to  present  the  subject  in  as  concise  a  form  as 
is  consistent  with  clearness,  to  make  it  attractive  and 
easily  intelligible  to  the  student,  and  at  the  same 
time  to  present  the  fullest  course  of  Trigonometry 
which  is  usually  given  in  the  best  Technological 
Schools. 

Considerable  care  has  been  taken  to  instruct  the 
student  in  the  theory  and  use  of  Logarithms,  and 
their  practical  application  to  the  solution  of  triangles. 
It  is  hoped  that  the  work  may  commend  itself,  not 
only  to  those  who  wish  to  confine  themselves  to  the 
numerical  calculations  which  occur  in  Trigonometry, 
but  also  to  those  who  intend  to  pursue  the  study  of 
the  higher  mathematics. 

The  examples  are  very  numerous  and  are  carefully 

selected.     Many  are   placed  in   immediate    connection 

with   the   subject-matter    which    they  illustrate.      The 

numerical   solution    of    triangles    has    received    much 

attention,   each   case   being    treated    in    detail.      The 

iii 


IV 


PREFACE. 


examples  at  the  ends  of  the  chapters  have  been  care- 
fully graded,  beginning  with  those  which  are  easy, 
and  extending  to  Jiose  which  are  more  and  more  diifi- 
cult.  These  examples  illustrate  every  part  of  the  sub- 
ject, and  are  intended  to  test,  not  only  the  student's 
knowledge  of  the  usual  methods  of  computation,  but 
his  ability  to  grasp  them  in  the  many  forms  they  may 
assume  in  practical  applications.  Among  these  exam- 
ples are  some  of  the  most  elegant  theorems  in  Plane 
and  Spherical  Trigonometry. 

The  Chapters  on  De  Moivre's  Theorem,  and  Astron- 
omy, Geodesy,  and  Polyedrons,  will  serve  to  introduce 
the  student  to  some  of  the  higher  applications  of 
Trigonometry,  rarely  found  in  American  textr-books. 

In  writing  this  book,  the  best  English  and  French 
authors  have  been  consulted.  I  am  indebted  especially 
to  the  works  of  Todhunter,  Casey,  Lock,  Hobson, 
Clarke,  Eustis,  Snowball,  M'Clelland  and  Preston, 
Smith,  and  Serret. 

It  remains  for  me  to  express  my  thanks  to  my  col- 
leagues. Prof.  R.  W.  Prentiss  for  reading  the  MS., 
and  Mr.  I.  S.  Upson  for  reading  the  proof-sheets. 

Any  corrections  or  suggestions,  either  in  the  text 
or  the  examples,  will  be  thankfully  received. 


E.  A.  B. 


RuTGEUs  College, 
New  Brunswick,  N.  J.,  April,  1892. 


TABLE     OF     CONTENTS. 


-ooj»;c 


PART  I. 


PLANE  TRIGONOMETRY. 


CHAPTER   I. 
Measurement  op  Angles. 

ART.  TkQ^ 

1.  Trigonometry 1 

2.  The  Measure  of  a  Quantity 1 

3.  Angles 2 

4.  Positive  and  Negative  Angles 3 

5.  The  Measure  of  Angles 3 

6.  The  Sexagesimal  Method 5 

7.  The  Centesimal  or  Decimal  Method 6 

8.  The  Circular  Measure 0 

9.  Comparison  of  the  Sexagesimal  and  Centesimal  Measures  . .  8 

10.  Comparison  of  the  Sexagesimal  and  Circular  Measures 9 

11.  General  Measure  of  an  Angle il 

12.  Complement  and  Supplement  of  an  Angle 12 

Examples 13 


CHAPTER   II.  ' 
The  Trigonometric  Functions. 

13.  Definitions  of  the  Trigonometric  Functions 16 

14.  The  Functions  are  always  the  Same  for  the  Same  Angle 18 

15.  Functions  of  Complemental  Angles 20 

16.  Representation  of  the  Functions  by  Straight  Lines 20 

17.  Positive  and  Negative  Lines 23 

V 


VI 


CO  J^  TENTS. 


ART.  PAQK 

18.  Functions  of  Angles  of  Any  Magnitude 23 

19.  Changes  in  Sine  as  tlie  Angle  increases  from  0°  to  .'JOO^ 26 

20.  Changes  in  Cosine  as  the  Angle  increases  from  0"  to  360°, . .  26 

21.  Changes  in  Tangent  as  the  Angle  increases  from  0°  to  360°.  27 

22.  Table  giving  Changes  of  Functions  in  Four  Quadrants 28 

23.  Relations  between  the  Functions  of  the  Same  Angle 29 

24.  Use  of  the  Preceding  Fonnulae 30 

25.  Graphic  Method  of  finding  the  Functions  in  Terms  of  One . .  30 

26.  To  find  the  Trigonometric  Functions  of  45° 31 

27.  To  find  the  Trigonometric  Functions  of  60°  and  30° 31 

28.  Reduction  of  Functions  to  1st  Quadrant 33 

29.  Functions  of  Coraplemental  Angles 34 

30.  Functions  of  Supplemental  Angles 34 

31.  To  prove  sin  (90°  +  A)  =  cos  A,  etc 35 

32.  To  prove  sin  (180°  +  A)  =  -  sin  A,  etc 35 

33.  To  prove  sin  (  —  A)  =^  —  sin  A,  etc 36 

34.  To  prove  sin  (270°  +  A)  =  sin  (270°  -  A)  =  -  cos  A,  etc. ...  36 
36.  Table  giving  the  Reduced  Functions  of  Any  Angle 37 

36.  Periodicity  of  the  Trigonometric  Functions 38 

37.  Angles  corresponding  to  Given  Functions 39 

38.  General  Expression  for  All  Angles  with  a  Given  Sine 40 

39.  An  Expression  for  All  Angles  with  a  Given  Cosine 41 

40.  An  Expression  for  All  Angles  with  a  Given  Tangent 41 

41.  Trigonometric  Identities 43 

Examples 44 


CHAPTER   III. 


Trigonometric  Functions  of  Two  Angles. 


42.  Fundamental  Formulae 50 

43.  To  find  the  Values  of  sin  (x  +  y)  and  cos  (x  +  y) 50 

44.  To  find  the  Values  of  sin  (x  —  y)  and  cos  (x  —  y) 52 

45.  Formulae  for  transforming  Sums  into  Products 65 

46.  Useful  FormulsB 56 

47.  Tangent  of  Sum  and  Difference  of  Two  Angles 57 

48.  Formulae  for  the  Sum  of  Three  or  More  Angles 58 

49.  Functions  of  Double  Angles 60 

50.  Functions  of  3  x  in  Terms  of  the  Functions  of  a; 61 

61.  Functions  of  Half  an  Angle 63 

62.  Double  Values  of  Sine  and  Cosine  of  Half  an  Angle 63 


il 


PAGE 


50 

60 
62 
65 
56 


68 
60 
61 
63 
63 


i 

<il 
'i 


CONTENTS.  vii 


ART.  PAOB 

^•^                       53.  Quadruple  Values  of  Sine  and  Cosine  of  Half  an  Angle 65 

25  54.  Double  Value  of  Tangent  of  Half  an  Angle 66 

26  65.  Triple  Value  of  Sine  of  One-third  an  Angle 67 

27  i          5(j.  Find  the  Values  of  the  Functions  of  22^° ,  69 

28  J         (,7.  Find  the  Sine  and  Cosine  of  18° 69 

58.  Find  the  Sine  and  Cosine  of  30° 70 

59.  If  A  +  B  +  C  =  180°,  to  find  sin  A  +  sin  B  +  sin  C,  etc 70 

60.  Inverse  Trigonometric  Functions 72 

(51.  Table  of  Useful  ForniulsB 75 

Examples 77 


29 
30 
30 
31 
31 
33 

34                         I  "     ■'    ■     ■    "      •           '•■; 

34  I  CHAPTER   IV. 

35  I 

35  Logarithms  and  Logarithmic  Tables.  —  Trigonometric  Tables. 

36  fl  62.   Nature  and  Use  of  Logarithms 87 

37  9  63. ,  Properties  of  Logarithms 87 

38  1^  64.   Comir  m  System  of  Logarithms 91 

39  A  65.   Comparison  of  Two  Systems  of  Logarithms 93 

40  M  66.  Tables  of  Logarithms 95 

41  H  67.   Use  of  Tables  of  Logarithms  of  Numbers 98 

41  H  68.   To  find  the  Logarithm  of  a  Given  Number 99 

43  9           69.  To  find  the  Number  corresponding  to  a  (liven  Logarithm. . .   102 

44  H  69a.   Arithmetic  Complement 103 

70.  Use  of  Trigonometric  Tables 105 

71.  Use  of  Tables  of  Natural  Trigonometric  Functions 106 

72.  To  find  the  Sine  of  a  Given  Angle 106 

73.  To  find  the  Cosine  of  a  Given  Angle 106 

74.  To  find  the  Angle  whose  Sine  is  Given ? 108 

75.  To  find  the  Angle  whose  Cosine  is  Given 108 

76.  Use  of  Tables  of  Logarithmic  Trigonometric  Functions 110 

77.  To  find  the  Logarithmic  Sine  of  a  Given  Angle 112 

78.  To  find  the  Logarithmic  Cosine  of  a  Given  Angle 112 

79.  To  find  the  Angle  whose  Logarithmic  Sine  is  Given 114 


-_  ■  80.  To  find  the  Angle  whose  Logarithmic  Cosine  is  Given 115 


81.   Angles  near  the  Limits  of  the  Quadrant 116 

Examples 117 


VUl 


CONTENTS. 


CHAPTER   V. 
Solution  of  Trigonometric  Equations. 

ART.  PAOB 

82.  Trigonometric  Equations 1*20 

83.  To  solve  m  sin  4>  =  a,  m  cos  (/»  =  ft 128 

84.  To  solve  a  sin  ^  +  &  cos  0  =  c 129 

85.  To  solve  sin  («  -f  a;)  =  m  sin  a; 131 

80.   To  solve  tan  (a  +  x)  =  m  tan  x 132 

87.  To  solve  tan  (a  +  x)  tan  x  =  ?m 133 

88.  To  solve  m  sin  (d  +  x)=  a,  m  sin  ((p  +  x)—.  h 134 

89.  To  solve  x  cos  a  +  y  sin  a  =  «t,  x  sin  u  —  y  cos  «  =  n 135 

90.  Adaptation  to  Logarithmic  Computation ■ 135 

91.  To  solve  7'C0S(/>  cos^  =  a,  rcos0  sin  0  =  b,  rsiui^  =  c 137 

92.  Trigonometric  Elimination 138 

Examples 140 

CHAPTER   VI. 

Relations  between  the  Sides  of  a  Triangle  and  the  Functions 

OF  its  Angles. 

93.  FormuliB 146 

94.  Right  Triangles 140 

95.  Oblique  Triangles  —  Law  of  Sines 147 

90.   Law  of  Cosines 148 

97.  Law  of  Tangents 149 

98.  To  prove  c  =  a  cos  B  +  ?>  cos  A 149 

99.  Functions  of  Half  an  Angle  in  Terms  of  the  Sides 150 

100.  To  express  the  Sine  of  an  Angle  in  Terms  of  the  Sides 152 

101.  Expressions  for  the  Area  of  a  Triangle 153 

102.  Inscribed  Circle '. 154 

103.  Circumscribed  Circle 154 

104.  Escribed  Circle 165 

105.  Distance  between  the  In-centre  and  the  Circumcentre 155 

106.  To  find  the  Area  of  a  Cyclic  Quadrilateral 157 

Examples 169 

CHAPTER  VII. 

Solution  op  Triangles. 

107.  Definitions 165 

108.  Four  Cases  of  Right  Triangles 165 


CONTENTS. 


IZ 


ART. 

109. 

110. 

111. 

PAOK 

....  12(5 

112. 

....  128 

11.3. 

....  129 

114. 

....  131 

115. 

....  132 

no. 

....  133 

117. 

...  1.34 

118. 

...  135 

119. 

...  135 

120. 

...  137 

1   I'-il- 

...  1.38 

:':'    122. 

...  140 

123. 

124. 

1       125. 

12(5. 

NCTIONS 

127. 

...  140 

...  146 

..147 

..  148 

( 

..  149 

128. 

..149     ! 

129. 

..150     i 

130. 

..152 

131. 

..  153     ' 

1.32. 

..154     : 

. .  154 

133. 

. .  155     l 

134. 

..155 

135. 

..157 

136. 

.  159 

137. 

165 


PAOI 

Case  I.  —  Given  a  Side  and  the  Ilypotenufie 1(5(5 

Ca.se  II.  — (iiven  an  Acute  Angle  and  the  llypotenuHe 1(57 

Case  III.  — Given  a  Side  and  an  Acute  Angle 168 

Case  IV.  —  Given  the  Two  Sides 160 

When  a  Side  and  the  Hypotenuse  are  nearly  Kcjual 1(59 

Four  Cases  of  Oblique  Triangles 172 

Case  I.  —  Given  a  Side  and  Two  Angles  172 

Case  II. — Given  Two  Sides  and  the  Angle  opposite  Gneof  them,  173 

Case  III.  —  Given  Two  Sides  and  the  Included  Angle 176 

Case  IV.  —  Given  the  Three  Sides 177 

Area  of  a  Triangle 180 

Heights  and  Distances  —  Definitions 181 

Heights  of  an  Accessible  Object 182 

Height  and  Distance  of  an  Inaccessible  Object 182 

An  Inaccessible  Object  above  a  Horizontal  Plane 184 

Object  observed  from  Two  Points  in  Same  Vertical  Line ....  185 

Distance  between  Two  Inaccessible  Objects 186 

The  Dip  of  the  Horizon 186 

Problem  of  Pothenot  or  of  Snellius 188 

Examples 189 


CHAPTER   VIII. 

Construction  of  Looauithmic  and  Tkioonometric  Tables. 

Logarithmic  and  Trigonometric  Tables 204 

Exponential  Series 204 

Logarithmic  Series 206 

Computation  of  Logarithms 207 

Sind  and  tan  6  are  in  Ascending  Order  of  Magnitude 208 

The  Limit  of  -—  is  Unity 209 

e 

Limiting  Values  of  sin  6  and  cos  d 209 

To  calculate  the  Sine  and  Cosine  of  10"  and  of  1' 211 

To  construct  a  Table  of  Natural  Sines  and  Cosines 213 

Another  Method 213 

138.  The  Sines  and  Cosines  from  30°  to  60" 214 

139.  Sines  of  Angles  Greater  than  45° 216 

140.  Tables  of  Tangents  and  Secants 216 

141.  Formulae  of  Verification 216 

142.  Tables  of  Logarithmic  Trigonometric  Functions 217 

143.  The  Principle  of  Proportional  Parts 218 


X  CONTENTS. 

ART.  PAGK 

144.  To  prove  the  Rule  for  the  Table  of  Common  Logarithms  . . .  218 

145.  To  prove  the  Rule  for  the  Table  of  Natural  Sines 219 

146.  To  prove  the  Rule  for  a  Table  of  Natural  Cosines 219 

147.  To  prove  the  Rule  for  a  Table  of  Natural  Tangents 220 

148.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Sines. .' 221 

149.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Cosines 222 

150.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Tangents 222 

151.  Cases  of  Inapplicability  of  Rule  of  Proportional  Parts 223 

152.  Three  Methods  to  replace  the  Rule  of  Proportional  Parts . . .  224 
Examples 226 


CHAPTER  IX. 
De  Moivre's  Theorem.  — Applications. 

163.  De  Moivre's  Theorem 229 

p 

164.  To  find  all  the  Values  of  (cos  e  +  V-l  sin  ey 231 

155.    To  develop  cos  nd  and  sin  7i6  in  Powers  of  sin  6  and  cos  0. . . .  233 

166.  To  develop  sin  d  and  cos  6  in  Series  of  Powers  ot  d 234 

167.  Convergence  of  the  Series 235 

168.  Expansion  of  cos"  6  in  Terms  of  Cosines  of  Multiples  of  ^. . .  235 

159.  Expansion  of  Bin"0  in  Terms  of  Cosines  of  Multiples  of  d. . .  236 

160.  Expansion  of  sin"  6  in  Terms  of  Sines  of  Multiples  of  ^ 237 

161.  Exponential  Values  of  Sine  and  Cosine 238 

162.  Gregory's  Series 239 

163.  Euler's  Series 240 

164.  Machin's  Series 241 

166.  Given  sin0  =  x  sin  (^  +  a)  ;  expand  0  in  Powers  of  a; 242 

166.  Given  tan  x  =  n  tan  e  ;  expand  x  in  Powers  of  n 242 

167.  Resolve  x"  —  1  into  Factors 243 

168.  Resolve  x"  +  1  into  Factors 244 

169.  Resolve  x""  -  2  r"  cos  5  +  1  into  Factors 245 

170.  De  Moivre's  Property  of  the  Circle 247 

171.  Cote's  Properties  of  the  Circle 248 

172.  Resolve  sin  6  into  Factors 248 

173.  Resolve  cos  $  into  Factors 260 

174.  Sum  the  Series  sin  .'  +  sin(  a+  j8)  +  etc 261 

175.  Sum  the  Series  ce    ^,  +  cos(rt  +  $)+  etc 252 

176.  Sum  the  Series  sin™  a  +  sin'"(rt  +  )9)  +  etc 252 

177.  Sum  the  Series  sin  «  -  sin(a  +  /3)  +  etc 254 

178.  Sum  the  Series  cosec e  +  cosec 2d  +  cosec 4 fl  +  etc 254 


i 


Tssfosm^sram 


COJf  TENTS. 


XI 


PAGE 

218 
219 
219 
220 
221 
222 
222 
223 
224 
226 


ABT.  PACK 

0  Q 

179.  Sum  the  Series  tan  9  +  ^  tan  -  +  \  tan  -  +  etc 256 

2  4 

180.  Sum  the  Series  sin  «  +  a;  sin  (  a+  3)  +  etc —  255 

181;   Summation  of  Infinite  Series  250 

Examples 267 


<jl»iOo~ 


PART  IL 
SPHERICAL  TRIGONOMETRY. 


229 


231    i 

■ 

233 

i   182. 

234 

1   ^^''^• 

235 

184. 

235    ^ 

185. 

236    1 

186. 

237 

187. 

238 

188. 

239 

189. 

240 

190. 

241 

191. 

242    s 

192. 

242    ' 

193. 

243 

194. 

244 

195. 

246 

196. 

247 

197. 

248 

198. 

248 

250 

261 

252 

262 

264 

199. 

264 

200. 

CHAPTER   X.  "\ 

Formula  relative  to  Spherical  Triangles. 

Spherical  Tnigonometry 267 

Geometric  Principles 267 

Fundamental  Definitions  and  Properties 268 

Formulje  for  Right  Spherical  Triangles 270 

Napier's  Rules 272 

The  Species  of  the  Parts 273 

Ambiguous  Solution 274 

Quadrantal  Triangles 274 

Law  of  Sines 276 

Law  of  Cosines 277 

Relation  between  a  Side  and  the  Three  Angles 278 

To  find  the  Value  of  cot  a  sin  6,  etc 279 

Useful  Formulae 280 

Formulae  for  the  Half  Angles 281 

Formulae  for  the  Half  Sides 284 

Napier's  Analogies 286 

DelambnVs  (or  Gauss's)  Analogies  287 

Examples 288 

.       CHAPTER   XL  /         .      , 

Solution  of  Si'heuical  Triangles. 

199.  Preliminary  Observations 297 

200.  Solution  of  Right  Spherical  Trii  agles 297 


BIIU..I.— .^1!-' 


Xll 


CONTENTS. 


ART.                                                                                       '  PAGK 

201.  Case  I.  —  Given  the  Hypotenuse  and  an  Angle 298 

202.  Case  II.  —  Given  the  Hypotenuse  and  a  Side 209 

203.  Case  III.  —  Given  a  Side  and  the  Adjacent  Angle 300 

204.  Case  IV.  —  Given  a  Side  and  the  Opposite  Angle ,301 

206.  Case  V.  —  Given  the  Two  Sides 302 

206.  Case  VI.  —  Given  the  Two  Angles 302 

207.  Quadrantal  and  Isosceles  Triangles 303 

208.  Solution  of  Oblique  Spherical  Triangles , . . .  304 

209.  Case  I.  —  Given  Two  Sides  and  the  Included  Angle 306 

210.  Case  II.  —  Given  Two  Angles  and  the  Included  Side 307 

211.  Case  III.  — Given  Two  Sides  and  One  Ci-posite  Angle.*. 309 

212.  Case  IV.  —  Given  Two  Angles  and  One  Opposite  Side 312 

213.  Case  V.  — Given  the  Three  Sides 313 

214.  Case  VI.  —  Given  the  Three  Angles 314 

Examples 316 


CHAPTER  XII. 
The  In-Circle8  and  Ex-Circles.  —  Areas. 

215.  The  Inscribed  Circle 324 

216.  The  Escribed  Circles 325 

217.  The  Circumscribed  Circle 326 

218.  Circumcircles  of  Colunar  Triangles 328 

219.  Areas  of  Triangles.  —  Given  the  Three  Angles 329 

220.  Areas  of  Triangles.  —  Given  the  Three  Sides 330 

221.  Areasof  Triangles. — Given  Two  Sides  and  the  Included  Angle,  331 
Examples 332 

CHAPTER  XIII. 
'Applications  of  Spherical  Trigonometry. 


222.  Astronomical  Definitions 338 

223.  Spherical  Coordinates 339 

224.  Graphic  Representation  of  the  Spherical  Coordinates 341 

225.  Problems 342 

226.  The  Chordal  Triangle 346 

227.  Legendre's  Theorem 348 

228.  Roy'.s  Rule .350 

229.  Reduction  of  an  Angle  to  the  Horizon 352 


324 
325 
326 
328 
329 
330 
igle,  331 
....  332 


338 
339 
341 
342 
346 
348 
360 
352 


CONTENTS. 


xm 


ART. 

FAQK 

230. 

298 

231. 

209 

232. 

300   j 

233. 

301 

234. 

302 

302 

303 

304 

305 

307  I 

309 

312   / 

313 

314 

316   ) 

Small  Variations  in  Parts  of  a  Spherical  Triangle 

Inclination  of  Adjacent  Faces  of  Polyedrons 

Volume  of  Parallelopiped 

Diagonal  of  a  Parallelopiped 

Table  of  Formulae  in  Spherical  Trigonometry 

Examples 


FAQK 

353 
,  356 
357 
358 
359 
,  362 


1 


TREATISE  ON  TRIGONOMETRY. 


3j*;«x^ 


PART  L 
PLANE    TRIGONOMETRY. 


CHAPTER  I.  ,     _ 

MEASUREMENT  OF  ANGLES. 

1.  Trigonometry  is  that  branch  of  mathematics  which 
treats  (1)  of  the  solution  of  plane  and  spherical  triangles, 
and  (2)  of  the  general  relations  of  angles  and  certain  func- 
tions of  them  called  the  trigonometric  functions. 

Plane  Tngonometry  comprises  the  solution  of  plane  trian- 
gles and  investigations  of  plane  angles  and  their  functions. 

Trigonometry  was  originally  the  science  which  treated  only  of  the 
sides  and  angles  of  plane  and  spherical  triangles ;  but  it  has  been 
recently  extended  so  as  to  include  the  analytic  treatment  of  all  theo- 
rems involving  the  consideration  of  angular  magnitudes. 

2.  The  Measure  of  a  Quantity.  —  All  measurements  of 
lines,  angles,  etc.,  are  made  in  terms  of  some  fixed  standard 
or  unit,  and  the  measure  of  a  quantity  is  the  number  of  times 
the  quantity  contains  the  unit. 

Tt  is  evident  that  the  same  quantity  will  be  represented 
by  different  numbers  when  different  units  are  adopted.  For 
example,  the  distance  of  a  mile  will  be  represented  by  the 
number  1  when  a  mile  is  the  unit  of  length,  by  the  number 
1760  when  a  yard  is  the  unit  of  length,  by  the  number  5280 
when  a  foot  is  the  unit  of  length,  and  so  on.     In  like  man- 


PLANE  TRIGONOMETRY. 


ner,  the  number  expressing  the  magnitude  of  an  angle  will 
depend  on  the  unit  of  angle. 

EXAMPLES. 

1.  What  is  the  measure  of  2^  miles  when  a  yard  is  the 

unit? 

2^  miles  =  |  X  1760  yards 

=  4400  yards  =  4400  x  1  yard. 
.*.  the  measure  is  4400  when  a  yard  is  the  unit. 

2.  What  is  the  measure  of  a  mile  when  a  chain  of  6a  feet 
is  the  unit  ?  Ans.  80. 

3.  What  is  the  measure  of  2  acres  when  a  square  whose 
side  is  22  yards  is  the  unit  ?  Aris.  20. 

4.  The  measure  of  a  certain  field  is  44  and  the  unit  is 
1100  square  yards ;  express  the  area  of  the  field  in  acres. 

Ans.  10  acres. 

5.  If  7  inches  be  taken  as  the  unit  of  length,  by  what 
.amber  will  15  feet  2  inches  be  represented  ?  Ans.  26. 

6.  If  192  square  inches  be  represented  by  the  number  12, 
what  is  the  unit  of  linear  measurement  ?        Ans.  4  inches. 

3.  Angles.  — An  angle  is  the  opening  between  two  straight 
lines  drawn  from  the  same  point.  The  point  is  called  the 
vertex  of  the  angle,  and  the  straight  lines  are  called  the  sides 
of  the  angle. 

An  angle  may  be  generated  by  revolving  a  line  from  coin- 
cidence with  another  line  about  a  fixed  point.  The  initial 
and  final  positions  of  the  line  are  the  ^3 
sides  of  the  angle;  the  amount  of 
revolution  measures  the  magnitude  of 
the  angle ;  and  the  angle  may  be  traced 
out  by  any  number  of  revolutions  of 
the  line.  O  A 

Thus,  to  form  the  angle  AOB,  OB  may  be  supposed  to 
have  revolved  from  OA  to  OB ;  and  it  is  obvious  that  OB 


THE  CIRCULAR  MEASURE. 


3 


may  go  on  revolving  until  it  comes  into  the  same  position 
OB  as  many  times  as  we  please ;  the  angle  AOB,  having  the 
same  bounding  lines  OA  and  OB,  may  therefore  be  greater 
than  2,  4,  8,  or  any  number  of  right  angles. 

The  line  OA  from  which  OB  moves  is  called  the  initial 
line,  and  OB  in  its  final  position,  the  terminal  line.  The 
revolving  line  OB  is  called  the  generatrix.  The  point  0  is 
called  the  origin,  vertex,  or  pole. 

4.  Positive  and  Negative  Angles.  — We  supposed  in  Art. 
3  that  OB  revolved  in  the  direction  opposite  to  that  of  the 
hands  of  a  watch.  But  angles  may,  of  course,  be  described 
by  a  line  revolving  in  the  same  direction  as  the  hands  of  a 
watch,  and  it  is  often  necessary  to  distinguish  between  the 
two  directions  in  which  angles  may  be  measured  from  the 
same  fixed  line.  This  is  conveniently  effected  by  adopting 
the  convention  that  angles  measured  in  one  direction  shall 
be  considered  positive,  and  angles  measured  in  the  opposite 
direction,  negative.  In  all  branches  of  mathematics  angles 
described  by  the  revolution  of  a  straight  line  in  the  direc- 
tion opposite  to  that  in  which  the  hands  of  a  watch  move 
are  usually  considered  positive,  and  all  angles  described  by 
the  revolution  of  a  straight  line  in  the  same  direction  as  the 
hands  of  a  watch  move  are  considered  negative. 

Thus,  the  revolving  line  OB  starts  from  the  initial  line 
OA.  When  it  revolves  in  the 
direction  contrary  to  that  of  the 
hands  of  a  watch,  and  comes  into 
the  position  OB,  it  traces  out  the 
positive  angle  AOB  (marked  +  a)  ; 
and  when  it  revolves  in  the  same 
direction  as  the  hands  of  a  watch, 
it  traces  the  negative  angle  AOB  (marked  —b). 

The  revolving  line  is  always  considered  negative. 

5.  The  Measure  of  Angles.  —  An  angle  is  measured  by 
the  arc  of  a  circle  whose  centre  is  at  the  vertex  of  the, 


-ip 


i 


-6 


4 


PLANE  TRIGONOMETRY. 


t  IE 


angle  and  whose  ends  are  on  the  sides  of  the  angle  (Geom., 

Art.  236). 

Let  the  line  OP  of  fixed  length  generate  an  angle  by 
revolving  in  the  positive  direction 
round  a  fixed  point  0  from  an  initial 
position  OA.  Since  OP  is  of  constant 
length,  the  point  P  will  trace  out  the 
circumference  ABA'B'  whose  centre 
is  0.  The  two  perpendicular  diam- 
eters AA'  and  BB'  of  this  circle  will 
inclose  the  four  right  angles  AOB, 
BOA',  A'OB',  and  B'OA. 

The  circumference  is  divided  at  the  points  A,  B,  A',  B' 
into  four  quadrants,  of  which 


AB  iQ  caXX^di  t\iQ  first  quadrant.  ■  • 

BA'  "       "       "     second  quadrant.  ' 

A'B'"       "       "     third  quadrant. 
J3r^  «       ti      ii    fourth  quadrant. 

In  the?  figure,  the  angle  AOPi,  between  the  initial  line 
OA  and  the  revolving  line  OPi,  is  less  than  a  right  angle, 
and  is  said  to  be  an  angle  in  the  first  quadrant.  AOP2  is 
greater  than  one  and  less  than  two  right  angles,  and  is  said  to 
be  an  angle  in  the  second  quadrant.  AOP3  is  greater  than  two 
and  less  than  three  right  angles,  and  is  said  to  be  an  angle 
in  the  third  quadrant.  AOP4  is  greater  than  three  and  less 
than  four  right  angles,  and  is  said  to  be  an  angle  in  the 
fourth  quadrant. 

When  the  revolving  line  returns  to  the  initial  position 
OA,  the  angle  AOA  is  an  angle  of  four  right  angles.  By 
supposing  OP  to  continue  revolving,  the  angle  described 
will  become  greater  than  an  angle  of  four  right  angles. 
Thus,  when  OP  coincides  with  the  lines  OB,  OA',  OB',  OA, 
in  the  second  revolution,  the  angles  described,  measured 
from  the  beginning  of  the  first  revolution,  are  angles  of  five 
right  angles,  six  right  angles,  seven  right  angles,  eight  right 


TUE  SEXAGESIMAL  METHOD.  6 

•  angles,  respectively,  and  so  on.  By  the  continued  revolu- 
tion of  OP  the  angle  between  the  initial  line  OA  and  the 
revolving  line  OF  may  become  of  any  magnitude  whatever. 

In  the  same  way  Or  may  revolve  in  the  negative  direc- 
tion about  0  any  number  of  times,  generating  a  negative 
angle  ;  and  this  negative  angle  may  obviously  have  any 
magnitude  whatever. 

The  angle  AOP  may  be  the  geometric  representative  of 
any  of  the  Trigonometric  angles  formed  by  any  number  of 
complete  revolutions,  either  in  the  positive  direction  added 
to  the  positive  angle  AOP,  or  in  the  negative  direction  added 
to  the  negative  angle  AOP.  In  all  cases  the  angle  is  said  to 
be  in  the  quadrant  indicated  by  its  terminal  line. 

Tliere  are  three  methods  of  measuring  angles^,  called 
respectively  the  Sexagesimal,  the  Centesimal,  and  the  Cir- 
cular methods. 


Isition 

By 

tribed 
ngles. 

i  OA, 
Isured 
If  five 
right 


I 


6.  The  Sexagesimal  Method.  —  This  is  the  method  in 
general  use.  In  this  method  the  right  angle  is  divided  into 
90  equal  parts,  each  of  which  is  called  a  degree.  Each 
degree  is  subdivided  into  60  equal  parts,  each  of  which  is 
called  a  minute.  Each  minute  is  subdivided  into  00  equal 
parts,  each  of  which  is  called  a  second.  Then  the  magni- 
tude of  an  angle  is  expressed  by  the  number  of  degrees, 
minutes,  and  seconds  which  it  contains.  Degrees,  minutes, 
and  seconds  are  denoted  respectively  by  the  symbols  °,  ',  " : 
thus,  to  represent  18  degrees,  6  minutes,  34.58  seconds,  we 

write 

'  18°  6'  34".58. 

A  degree  of  arc  is  -^\-q  of  the  circumference  to  which  the 
arc  belongs.  The  degree  of  arc  is  subdivided  in  the  same 
manner  as  the  degree  of  angle. 

Then      1  circumference  =  3G0°  =  21600'  =  1296000". 
1  quadrant  or  right  angle  =  90°. 

Instruments  used  for  measuring  angles  are  subdivided 
accordingly. 


6 


PLANE  THIGONOMETRY. 


7.  The  Centesimal  or  Decimal  Method.  —  In  this  method 
the  right  angle  is  divided  into  100  equal  parts,  each  of 
which  is  called  a  grade.  Each  grade  is  subdivided  into  100 
equal  parts,  each  of  which  is  called  a  minute.  Each  minute 
is  subdivided  into  100  equal  parts,  each  of  which  is  called 
a  second.  The  magnitude  of  an  angle  is  then  expressed  by 
the  number  of  grades,  minutes,  and  seconds  which  it  con- 
tains. Grades,  minutes,  and  seconds  are  denoted  respect- 
ively by  the  symbols  %  \  '''* :  thus,  to  represent  34  grades, 
48  minutes,  8G.47  seconds,  we  write 

34«  48^  86^\47. 

The  centesimal  or  decimal  method  wan  proposed  by  the  French 
mathematicians  in  the  beginning  of  the  present  century.  But 
altliougli  it  possesses  many  advantages  over  the  established  method, 
tliey  were  not  considered  sufficient  to  counterbalance  the  enormous 
labor  which  would  have  been  necessary  to  rearrange  all  the  mathe- 
matical tables,  books  of  reference,  and  records  of  observations,  which 
would  have  to  be  transferred  into  the  decimal  system  before  its 
advantages  could  be  felt.  Thus,  the  centesimal  method  has  never  been 
used  even  in  France,  and  in  all  probability  never  will  be  used  in  prac- 
tical work. 


8.  The  Circular  Measure.  —  The  unit  of 
circular  measure  is  the  angle  subtended  at 
the  centre  of  a  circle  by  an  arc  equal  in 
length  to  the  radius. 

This  unit  of  circular  measure  is  called  a 
radian. 

Let  0  be  the  centre  of  a  circle  whose  radius  is  r 

Let  the  arc  AB  be  equal  to  the  radius 
OA  =  r. 

Then,  since  angles  at  the  centre  of  a 
circle  are  in  the  same  ratio  as  their 
intercepted  arcs  (Geom.,  Art.  234),  and 
since  the  ratio  of  the  circumference  of 
a  circle  to  its  diameter  is  tt  =  3.14159265 
(Geom.,  Art.  436), 


THE  CIRCULAR   MEASURE. 


.'.  angle  AOB  :  4rt.  angles  : :  arc  AB  :  circumi'erence, 

:  :  r  :  2irr  : :  1  :  'Jv. 


an 


gle  AOB  =  ^  ^'^-  ^"g^^s  ^  2  I't.  angles  ^ 

27r  TT 

180° 


a  radian  =  ansrle  AOB  = 

"  3.14159265 


=  57°.2957795 


=  3437'.74677  =  206264".80G. 

Therefore,  the  radian  is  the  same  for  all  circles,  and 
=  57°.2957795.  ^^ ^B 

Let  ABP  be  any  circle ;  let  the  angle 
AOB  be  the  radian;  and  let  AOP  be 
any  other  angle.  I  ^ — ^         |A 

Then  arc  AB  =  radius  OA. 

.•.  angle  AOP  :  angle  AOB 

: :  arc  AP  :  arc  AB  ; 

or      angle  AOP  :  radian  : :  arc  AP  :  radius.  < 

.-.  an  gle  AOP  =  —^; — x  radian, 
radius 

The  measure  of  any  quantity  is  the  number  of  times  it 
contains  the  unit  of  measure  (Art.  2). 

.'.  the  circular  measure  of  angle  AOP  = 

radius 

Note  1.  —The  student  will  notice  that  a  radian  ia  a  little  less  than  an  angle  of  an 
equilateral  triangle,  i.e.,  of  60°. 

Angles  expressed  in  circular  measure  are  usually  denoted  by  Qreek  letters,  a,  p, 
y,  .. .,<!>,  0,  ^ 

The  circular  measure  is  employed  in  the  various  branches  of  Analytical  Mathe- 
matics, in  which  the  angle  under  consideration  is  almost  always  expressed  by  » 
letter. 

Note  2.  —  The  student  cannot  too  carefully  notice  that  unless  an  angle  is  obvi- 
ously  referred  to,  the  letters  a,  /3, ...,  0,  <j), ...  stand  for  mere  nuvibers.  Thus,  rr  stands 
for  a  number,  and  a  number  only,  viz.,  3.14159  ...,  but  in  the  expression  '  the  angle 
IT,'  that  is, '  the  angle  3.14159  ...,'  there  must  be  some  unit  understood.  The  unit 
understood  here  is  a  radian,  and  therefore  '  the  angle  it  '  stunds  for  '  n  radians'  or 
3.14159  ...  radians,  that  is,  two  right  angles. 

Hence,  when  an  angle  is  referred  to,  n  is  a  very  convenient  abbreviation  for 
two  right  angles. 

So  also  '  the  angle  a  or  0 '  means  !  a  radians  or  9  radians.' 

The  units  in  the  three  systems,  when  expressed  in  terms 
of  one  common  standard,  two  right  angles,  stand  thus  : 


8 


PLANE  THIGONOMETUY. 


■\ 


The  unit  in  the  Sexagesimal  Method  =  — —  of  2  right  angles. 


u 


u 


(t 

"    "    Centesimal 

i( 

200 

(I 

(t 

"    "    Cirjular 

« 

=    -    "  " 

« 

(( 


« 


ir 


If  D,  G,  and  $  denote  the  number  of  degrees,  grades,  and 
radians  respectively  in  any  angle,  then 

180     200     tt'  '^  ^ 

because  each  fraction  is  the  ratio  of  the  angle  to  two  right 
angles. 

9.  Comparison  of  the  Sexagesimal  and  Centesimal  Meas- 
ures of  an  Angle.  —  Although  the  centesimal  method  was 
never  in  general  use  among  mathematicians,  and  is  now 
totally  abandoned  everywhere,  yet  it  still  possesses  some 
interest,  as  it  shows  the  application  of  the  decimal  system 
to  the  measurement  of  angle?. 

From  (1)  of  Art.  8  we  have  , 

_D    ^_G  \     .'       ^     ;'    "  ;.. 

180      200* 


.-.  D  =  :^G,  andG=^D. 


m     '[; 


EXAMPLES. 

1.    Express  49°  15'  35"  in  centesimal  measure. 
First  express  the  angle  in   degrees  and  decimals  of  a 
degree  thus : 

60)  35" 

60)  15' .58.3 

49°.25972 
10 

9)  492.5972 

54K.733024.... 

.-.  49°  15'  35"=  548  j^^  30^\24  .... 


COMPARISON  OF  MEASUHE8. 


9 

2.   Express  87«  2^  25^^  in  degrees,  etc 
^^  First  express  tlie  angle  in  grades  and  decimals  of  a  grade 

87«2^25^^=87<f.0225 

•     ^_^ .9_ 

,      78.320:^5 
,     60 

:   ^  19.215 

,'■■  ■'.'■'',  ■ CO 

12.9 

'  .-.  87*^2^  25^^=  78M9'12".9. 

Find  the  number  of  grades,  minutes,  and  seconds  in  the 
lollo wing  angles : 

^-  ^1°    4|30".  Am.  mis'   0«. 

4.  45°  33'    3".                                        60*6r20"37 

5-  27°  15' 40".                                        30«29M9«.75'... 

«•  1«^°    *'    9"-                                      174-52M2«.962..'.. 

Find  the  number  of  degrees,  minutes,  and  seconds  in  the 
loilowing  angles: 

7.     19.4^96".  ^,«.  17°  30' 48"  78 

8-   124«   5^    8"..  .       .  111°  38' 44".692. 

a     65«18^35«.  49°  39' 54".54. 

10.   Comparison  of  the  Sexagesimal  and  Circular  Meas- 
ures  of  an  Angle. 
From  (1)  of  Art.  8  we  have  ^     >  ' 

■.,■;■■,:■  180     TT  ,;^ '■..'",:. 

'   .M)  =  l^and^  =  ^. 

7^  180  :     .     . 


10 


PLANE  TRIGONOMETRY. 


m 


I 
I 


EXAMPLES. 

1.   Find  the  number  of  degrees  in  the  angle  whose  circu- 

1  '     '    ■-■ 


lar  measure  is  | 


Here 


e  = 


2 


TT  2 


TT 

90x7 

22  " 


=  28°  38'  10"iA 


where  ^  is  used  for  TT. 

2.  Find  the  circular  measure  of  the  angle  59°  52'  30".     • 

Express  the  angle  in  degrees  and  decimals  of  a  degree 

thus:  :  •• 

60)52.5 

:  59.875 

...  ^=5i^_p^  =  (.333...)  7r=-«  .0453..-. 

3.  Express,  in  degrees,  the  angles  whose  circular  measures 

TT       TT        TT        TT         2 

2'    3'    4'    6'    ^ 

NoTB  1.— The  student  should  ospeoially  accustom  himself  to  express  readily  iu 
ciruular  measure  ud  uugle  which  is  given  ia  d'^grees. 


4.   Express  in  circular  measure  the  following  angles 


60°,  22°  30',  11°  15',  270°. 


Ans. 


IT        TT         IT 


3: 


3'    8'    16'     2 

5.   Express  in  circular  measure  3°  12',  and  find  to  seconds 
the  angle  whose  circular  measure  is  .8.  ,.      ;    . 

4: 

T 


/^TakeTT^—.' 


Ans.  — ,    45°  49'  5"fV. 
225  ^^ 


6.    One  angle  of  a  triangle  is  45°,  and  the  circular  measure 
of  another  is  1.5.     Find  the  third  angle  in  degrees. 

Ans.  49°  5'  27"^^. 

Note  2.  —  Questions  in  which  angles  are  expressed  in  different  systems  of  meas- 
urement are  easily  solved  by  expressing  each  angle  in  right  angles. 


.-_^i 


GENERAL  MEASURE  OF  AN  ANGLE. 


11 


7.  The  sum  of  the  measure  of  au  angle  in  degrees  and 
twice  its  measure  in  radians  is  23^;  find  its  measure  in 
degrees  (tt  =  ^y? ) . 

Let  the  angle  contain  x  right  angles.      ''  "  ^" 

Then  the  measure  of  the  angle  in  degrees  =  90  x. 


it 


t( 


l( 

ic      a 

((       a 

radians  = 

X. 

.:  90a; 

+  «■« 

=  23^; 

.-.  90  a: 

+  v-^ 

163 

~   7  ' 

.-.  652a; 

=  163,  .-. 

a;: 

_1 

4 

the 

angle  is 

i  of  90° 

=  221°. 

8.  The  difference  between  two  angles  is  -,  and  tlieir  sum 
is  56° ;  find  the  angles  in  degrees.  Arts.  38°,  18°. 

11.  General  Measure  of  an  Angle.  —  In  Euclidian  geom- 
etry and  in  practical  applications  of  trigonometry,  angles 
are  generally  considered  to  be  less  than  two  right  angles  ; 
but  in  the  theoretical  parts  of  mathematics,  angles  are 
treated  as  quantities  which  may  be  of  any  magnitude  what- 
ever. 

Thus,  when  we  are  told  that  an  angle  is  in  some  particu- 
lar quadrant,  say  the  second  (Art.  5),  we  kuow  that  the 
position  in  which  the  revolving  line  stops  is  in  the  second 
quadrant.  But  there  is  an  unlimited  number  of  angles 
having  the  same  final  position,  OP. 

The  revolving  line  OP  may  pass  from 
OA  to  OP,  not  only  by  describing  the 
arc  ABP,  but  by  moving  through  a  whole 
revolution  plus  the  arc  ABP,  or  through 
any  number  of  revolutions  plus  the  aro 
ABP. 

For  example,  the  final  position  of  OP 
may  represent  geometrically  all  the  fol- 
lowing angles : 


12 


PLANE   TRIGONOMETRY. 


Angle  AOP  =  130°,  or  360° +  130°,  or  720° +  130°,  or 
-  360°  +  130°,  or  -  720°  +  130°,  etc. 

Let  A  be  an  angle  between  0  and  90°,  and  let  n  be  any 
lohole  number,  positive  or  negative.     Then 

(1)  271  X  180°  +  A  represents  algebraically  an  angle  in  the 

first  quadrant. 

(2)  2w  X  180°  —  A  represents  algebraically  an  angle  in  the 

/owrf/i  quadrant.  " 

(3)  (2n  +  1)  180°  —  A  represents  algebraically  an  angle  in 

the  second  quadrant. 

(4)  (2  n  +  1 )  180°  +  A  represents  algebraically  an  angle  in 

the  f/iiVd  quadrant.  .  .     , 

In  circular  measure  the  corresponding  expressions  are 
(1)  2nir+d,  (2)  2mr-e,  (3)  {2n  +  l)TT-6,  (4)  (2n  +  l)7r+^. 


EXAMPLES.  ..  ■!•  • 

State  in  which  quadrant  the  revolving  line  will  be  after 
d??scribing  the  following  angles  : 

(1)  120°,    (2)  340°,    (3)  490°,    (4)  -  100°, 

(5)  _  380°,    (6)  Itt,    (7)  IOtt  + 


IT 


12.  Complement  and  Supplement  of  an  Angle  or  Arc.  — 

The  complement  of  an  angle  or  arc  is  the  remainder  obtained 
by  subtracting  it  from  a  right  angle  or  90°. 

The  supplement  of  an  angle  or  arc  is  the  remainder 
obtained  by  subtracting  it  from  tivo  right  angles  or  180°. 

Thus,     the  complement  of  A  is  (90°  —  A). 

The  complement  of  190°  is  (90°  -  190°)  =  -100°. 

The  supplement  of  A  is  (180°  -  A). 

The  supplement  of  200°  is  (180°  -  200°)  =  -  20°. 

The  complement  of  f  tt  is  ( ^  —  ^tt  j  = 
The  supplement  of  \ir  is  {j  —  \Tr)=\ir. 


|t- 


m\ 


1 ' 


EXAMPLES. 


13 


or 


EXAMPLES. 

1.  If  192  square  inches  be  represented  by  the  number  12, 
what  is  the  unit  of  linear  measurement  ?         Ans.  4  inches. 

2.  If  1000  square  inches  be  represented  by  the  number 
40,  what  is  the  unit  of  linear  measurement  ?    Ans.  5  inches. 

3.  If  2000  cubic  inches  be  represented  by  the  number  16, 
what  is  the  unit  of  linear  measurement  ?        Ans.  5  inches. 

4.  The  length  of  an  Atlantic  cable  is  2300  miles  and  the 
length  of  the  cable  from  England  to  France  is  21  miles. 
Express  the  length  of  the  first  in  terms  of  the  second  as 
unit.  Ans.  lOO^. 

5.  Find  the  measure  of  a  miles  when  b  yards  is  the  unit. 

.        1760a 
Ans. 

b 

6.  The  ratio  of  the  area  of  one  field  to  that  of  another  is 

20  :  1,  and  the  area  of  the  first  is  half  a  square  mile.     Find 
the  number  of  square  yards  in  the  second.  Ans.  77440. 

7.  A  certain  weight  is  3.125  tons.  What  is  its  measure 
in  terms  of  4  cwt.?  Ans.  15.625. 

Express  the  following  12  angles  in  centesimal  measure : 


' 

8. 

42°  15'  18". 

Ans.  m  95\ 

9. 

63°  19'  17". 

70B35^70^\98.... 

ned 

10. 

103°  15'  45". 

1148  73V  g^vv  1 

11. 

19°    0'18". 

218  UN  66^\6. 

ider 

12. 

143°    9'    0". 

1598    5^55^5.      V 

13. 

300°  15'  58". 

3338  62^  90^\l234567896. 

14. 

27°  41'  51". 

308.775. 

00°. 

15. 

67°.4325. 

748.925. 

', .. 

16. 

8°  15'  27". 

98l7^50^ 

20°. 

17. 

97°    5' 15". 

1078  87^  50^\ 

18. 

16°  14'  19". 

188    4^29^^.... 

19. 

132°    G'. 

1468  7r77^\7. 

14 


PLANE  TRIGONOMETRY 


Express  the  following  11  angles  in  degrees,  minutes,  and 
seconds : 


20.  105K  52^  75^\ 

21.  82«    9^54^ 

22.  70K  15^  92^\ 

23.  158    0^15^ 

24.  1548    7N24^\ 

25.  324«13^88^7. 

26.  lOe  42^  50^\ 

27.  208  77^50^ 

28.  8«75\ 

29.  170M5^35^ 

30.  248    o^25^\ 


Ans.  94°  58'  29".l. 
73°  53'    9".096. 
63°    8'35".808. 
13°  30'    4".86. 
138°  39'  54".576. 
291°  43'  29".9388. 
9°  22' 57". 
18°  41' 51". 
7°  52' 30".       ■. 
153°  24' 29".34. 
■  21°  36'    8".l.  ' 


Express  in  circular  measure  the  following  angles  : 

•        '  14537r 


31.  315°,  24°  13'. 

32.  95°  20',  12°  5'  4". 

33.  221°,  r,  57°.295. 

34.  120°,  45°,  270°. 

35.  360°,  3^-  rt.  angles. 


Ans.  ^TT, 


10800 
143  TT    2719  TT 
270  '    40500* 


TT 


TT 


8'    180' 


1  radian. 


ir 


2.09439,  |,    f  TT. 

27r,    Itt. 


Express  in  degrees,  etc.,  the  angles  whose  circular  meas- 
ures are :  ■ 


36. 


f '^'    I'r'    9* 


07    1    1    2 
^^'   4'  6'  3' 


5 


38.   -,  .7854. 
6 


Ans.  112°.5,  120°,  ??  degrees. 

45,  30,  120, 

—  degrees,  —  degrees,   degrees. 

.TT  TT  TT 

47°43'38"3^,  45°. 


il 


EXAMPLES. 


15 


39.   ^,  j\w,  2.504.     A,is.  257°  49'  43".39,  15°,  143°.468. 


40.   .0234,  1.234,  |. 
o 


1°20'27",  70°  42' 11",  38°  11' 50" 


41.  Find  the  number  of  radians  in  an  angle  at  the  centre 
of  a  circle  of  radius  25  feet,  which  intercepts  an  arc  of 
37^  feet.  Ans.  1^. 

42.  Find  the  number  of  degrees  in  an  angle  at  the  centre 
of  a  circle  of  radius  10  feet,  which  intercepts  an  arc  of 
57rfeet.  ■  ■  ■  '  -       '  ■  Ans.  90°. 

43.  Find  the  number  of  right  angles  in  an  angle  at  the 
centre  of  a  circle  of  radius  3^  inches,  which  intercepts  an 
arc  of  2  feet.  .  ,    j^jig^  4^^ 

44.  Find  the  length  of  the  arc  subtending  an  angle  of 
i^  radians  at  the  centre  of  a  circle  whose  radius  is  25  feet. 

,    ^  Ans.  112|ft. 

45.  Find  the  length  of  an  arc  of  80°  on  a  circle  of  4  feet 
radius.  Ans.  5f^  ft. 

46.  The  angle  subtended  by  the  diameter  of  the  Sun  at 
the  eye  of  an  observer  is  32' :  find  approximately  the 
diameter  of  the  Sun  if  its  distance  from  the  observer  be 
90  000  000  miles.  Ans.  838  000  miles. 

47.  A  railway  train  is  travelling  on  a  curve  of  half  a  mile 
radius  at  the  rate  of  20  miles  an  hour :  through  what  angle 
has  it  turned  in  10  seconds  ?  Ans.  6^  degrees. 

48.  If  the  radius  of  a  circle  be  4000  miles,  find  the  length 
of  an  arc  w^ich  subtends  an  angle  of  1"  at  the  centre  of  the 
circle.  Ans.  About  34  yards. 

49.  On  a  circle  of  80  feet  radius  it  was  found  that  an 
angle  of  22°  30'  at  the  centre  was  subtended  by  an  arc 
31  ft.  5  in.  in  length :  hence  calculate  to  four  decimal  places 
the  numerical  value  of  the  ratio  of  the  circumference  of  a 
circle  to  its  diameter.  Ans.  3.1416. 

50.  Find  the  number  of  radians  in  10"  correct  to  four  sig- 
nificant figures  (use  ff^f  for  tt).  Ans.  .00004848. 


16 


PLANE  TRIGONOMETRY. 


H 


CHAPTER   11. 

THE  TEIGONOMETEIO   FUNCTIONS. 

13.  Definitions  of  the  Trigonometric  Functio.ns.  —  Let 
EAD  be  an  angle;  in  AD,  one  of  the  lines  containing  the 
angle,  take  any  point  B,  and  from  B 
draw  BC  perpendicular  to  the  other 
line  AR,  thus  forming  a  right  triangle 
ABC,  right-angled  at  C.  Then  denot- 
ing the  angles  by  the  capital  letters  A, 
B,  C,  respectively,  and  the  three  sides 
opposite  these  angles  by  the  corresponding  small  italics,  a, 
h,  c,*  we  have  the  following  definitions : 

opposite  side 


a_ 

c       hypotenuse 
h  __  adjacent  side 
c       hypotenuse 


is  called  the  sine  of  the  angle  A. 
is  called  the  cosine  of  the  angle  A. 


-  =  ^f — ~  is  called  the  tangent  of  the  angle  A. 

0     adjacent  side 


h_  adjacent  side 
a     opposite  side 

c  _  hypotenuse 
b      adjacent  side 

c_  hypotenuse 
a     opposite  side 


is  called  the  cotangent  of  the  angle  A. 
is  called  the  secant  of  the  angle  A. 
is  called  the  cosecant  of  the  angle  A. 


If  the  cosine  of  A  be  subtracted  from  unity,  the  remain- 
der is  called  the  versed  sine  of  A.     If  the  sine  of  A  be  sub- 


*  The  letters  a,  b,  c  are  numbers,  being  the  number  of  times  the  lengths  of  the  sides 
contain  some  chosoo  unit  of  length. 


TRIGONOMETRIC  FUNCTIONS. 


17 


1 


A.       - 


tracted  from  unity,  the  remainder  is  called  the  coversed  sine 
of  A ;  the  latter  term  is  hardly  ever  used  in  practice. 

The  words  sine,  cosine,  etc.,  are  abbreviated,  and  the  func- 
tions of  an  angle  A  are  written  thus  :  sin  A,  cos  A,  tan  A, 
cot  A,  sec  A,  cosec  A,  vers  A,  covers  A. 

The  following  is  the  verbal  enunciation  of  these  defini- 
tions: 

The  sine  of  an  angle  is  the  ratio  of  the  oj^posite  side  to  the 

hypotenuse;  orsinA  =  -' 

c 

The  cosine  of  an  angle  is  the  ratio  of  the  adjacent  side  to 

the  hypotenuse ;  or  cos  A  =  —  ■  /' 

c 

The  tangent  of  an  angle  is  the  ratio  of  the  opposite  side  to 


the  adjacent  side;  or  tan  A  = 


a 


The  cotangent  of  an  angle  is  the  ratio  6f  the  adjacent  side  to 

b 


the  opposite  side;  or  cot  A  = 


a 


The  secant  of  an  angle  is  the  ratio  of  the  hypotenuse  to  the 

adjacent  side ;  orsecA  =  —  ..,.,'     .\v 

h       -:'"  ;;  ■•  ^ '     ■■  •■    ■■  ■" '  ■'  .'■■   '  ■'• 

The  cosecant  of  an  angle  is  the  ratio  of  the  hypotenuse  to  the 


opposite  side;  or  cosec  A  = 


a 


The  versed  sine  of  an  angle  is  unity  minus  the  cosine  of  the 


angle;  or  vers  A  =  1  —  cos  A  =  1 • 

c 


I      ii 


The  coversed  sine  of  an  angle  is  unity  mini,  i  the  sine  of  the 


a 


angle;  or  covers  A  =  1    -  sin  A  =  1 


These  ratios  are  called  Trigonometric  Functions.  The 
student  should  carefully  commit  them  to  memory,  as  upon 
them  is  founded  the  whole  theory  of  Trigonometry. 

These  functions  are,  it  will  be  observed,  not  lengths,  but 


18 


PLANE  TRIGONOMETRY. 


ratios  of  one  length  to  anotlier ;  that  is,  they  are  abstract 
numbers,  simply  numerical  quantities ;  and  they  remain 
unchanged  so  long  as  the  angle  remains  unchanged,  as  will 
be  proved  in  Art.  14. 

It  is  clear  from  the  above  definitions  that 


:M 


.n 


cosec  A 

X 

or 

sinA  = 

sin  A' 

sec  A 

1 

cos  a' 

or 

cosA  = 

tan  A 

1 
—     .  .  > 

or 

cot  A  = 

cot  a' 


cosec  A' 

1 
sec  a' 


tan  A 


The  powers  of  the  Trigonometric  functions  are  expressed 

as  follows; 

(sin  A)2  is  written  sin^  A, 
(cosA)^  is  written  cos^A,    >^>;  '  '  -> 

and  so  on. 

Note,  —  The  student  must  notice  that  •  sin  A  '  is  a  single  symbol,  the  name  of  a 
number,  or  fraction  belonging  to  the  angle  A.  Also  sin^A  is  an  abbreviation  for 
(sin  A)',  i.e.,  for  (sin  A) x  (sin  A).    Such  abbreviations  are  used  for  convenience. 

14.  The  Trigonometric  Functions  are  always  the  Same 
for  the  Same  Angle.  —  Let  BAD  be 

any  angle ;  in  AD  take  P,  P',  any 
two  points,  and  draw  PC,  P'C  per- 
pendicular to  AB.  Take  P",  any 
point  in  AB,  and  draw  P"C"  per- 
pendicular to  AD.  A  C  C  P 

Then  the  three  triangles  PAC,  P'AC,  P"AC"  are  equi- 
angular, since  they  are  right-angled,  and  have  a  common 
angle  at  A :  therefore  they  are  similar. 

"  AP      AP'"~  AP"* 

But  each  of  these  ratios  is  the  sine  of  the  angle  A.  Thus, 
sin  A  is  the  same  whatever  be  the  position  of  the  point  P  on 
either  of  the  lines  containing  the  angle  A. 


\ 


i: 


FUNCTIONS   OF  COMPLEMENTAL  ANGLES. 


19 


Therefore   sin  A  is  always  the  same.     A  similar  proof 
may  be  given  for  each  of  the  other  functions. 
In  the  right  triangle  of  Art.  13,  show  that 

a  =  c  sin  A  =  c  cos  B  =  &  tan  A  =  b  cot  B, 
b  —  a  cot  A  =  tt  tan  B  =  c  cos  A  =  c  sin  B, 
c  =  a  cosec  A  =  a  sec  B  =  6  sec  A  =  6  cosec  B. 

Note.  —These  results  should  be  carefully  noticed,  as  they  are  of  frequent  use  in 
the  solution  of  right  triangles  and  elsewhere. 

EXAMPLES. 

1.  Calculate  the  value  of  the  functions,  sine,  cosine,  etc., 
of  the  angle  A  in  the  right  triangles  whose  sides  a,  b,  c  are 
respectively  (1)  8, 15,  17;  (2)  40,  9,  41 ;  (3)  196,  315,  371 ; 
(4)  480,  31,  481 ;  (5)  1700,  945,  1945. 

Ans.  (1)  sin  A  =  -^-j,    cos  A  =  \^,  tan  A  =  -^-^f  etc. ; 

,    (2)  sin  A  =  1^,    cos  A  =  ^\,  etc. ; 

(3)  sinA  =  ff,    tanA  =  |f,  etc.; 

;  (4)  sin  A  =  f  ff ,  tan  A  =  -\\%  etc. ; 

(5)  sin  A  =  III,  tan  A  =  f|f,  etc. 

In  a  right  triangle,  given : 


2.   a  =  Vm^+  n%  b  =  V2  mn ;  calculate  sin  A. 

Ans. 


3.  a  —  -\/'m?—mn,  b  =  n\  calculate  sec  A. 

4.  a  =  Vw^+ mn,  c=m+n;  calculate  tan  A.     -y/ 

5.  a  =  2  miij  b  =  m^—n^;  calculate  cos  A. 


m  +  n 

m  —  n 


n 


mn  +n^ 


m" 


7J/ 


6.  sin  A  =  1^,  c  =  200.5  ;  calculate  a. 

7.  cos  A  =  .44,  c  =  30.5 ;  calculate  b. 

8.  tan  A  =  -y-,  b  =  ^;  calculate  c. 


120.3. 

13.42. 

T^rViSO. 


vj 


f 


20 


PLANE  TRIGONOMETRY. 


15.  Functions  of  Complemental  Angles.  —  In  the  rt.  A  ABC 

we  have 

siu  A  =  -,  and  cosB  =  -'•    (Art.  13.) 
c  c     \ 

.'.  sin  A  =  cosB. 

But  B  is  the  comphMuent  of  A,  since 
their  sum  is  aright  angle,  or  90°;  i.e.,  A 
B  =  90°-A. 

.-.  sinA       =cosB       =  cos  (90°— A)       = 


Also,       cos  A 


sinB       =  sin  (90°- A)       =  ", 


a 


tan  A       =cotB       =  cot  (90°- A)       = -, 

0 

cot  A       =tanB       =  tan  (90°- A)      =-, 

(a/ 

sec  A       =cosecB    =  cosec  (90°— A)   =^, 
cosec  A   =  secB       =  sec  (90°—  A)       =  -, 
vers  A     =  covers  B  =  covers  (90°  —  A)  =r  1  —  -, 
covers  A  =  vers  B     =  vers  (90°  —  A)     =  1  — 


Therefore  the  sine,  tangent,  secant,  and  versed  sine  of  an 
angle  are  equal  respectively  to  the  cosine,  cotangent,  cosecant, 
and  coversed  sine  of  the  comi^lement  of  the  angle.        * 

16.  Representation  of  the  Trigonometric  Functions  by 
Straight  Lines.  —  The  Trigonometric  functions  were  for- 
merly defined  as  being  certain  straight  lines  geometrically 
connected  with  the  arc  subtending  the  angle  at  the  centre 
of  a  circle  of  given  radius. 

Thus,  let  AP  be  the  arc  of  a  circle  subtending  the  angle 
AOP  at  the  centre. 


I 


REPRESENTATION  OF  FUNCTIONS  BY  LINES.      21 

Draw  the  tangents  AT,  BT'  meeting  01*  produced  to  T', 
and  draw  PC,  VD  ±  to  OA,  OB. 


Then 


« 

cotangent 

(( 

« 

secant 

(I 

((    - 

cosecant 

i( 

ft 

versed  sine 

(( 

(I   ■ , 

.     CO  versed  sine 

(I 

PC  was  called  the  sine  of  the  arc  AP. 

DC         ,     "  cosine  " 

AT  «  tangent  " 

BT' 

OT 

OT' 

AC 

BD 

*     ■ 

Since  any  arc  is  the  measure  of  the  angle  at  the  centre 
which  the  arc  subtends  (Art.  5),  the  above  functions  of  the 
arc  AP  are  also  functions  of  the  angle  AOP. 

It  should  be  noticed  that  the  old  functions  of  the  arc  above 
given,  when  divided  by  the  radius  of  the  circle,  become  the 
modern  functions  of  the  angle  which  the  arc  subtends  at  the 
centre.  If,  therefore,  the  radius  be  taken  as  unity,  the  old 
functions  of  the  arc  AP  become  the  modern  functions  of  the 
angle  AOP. 

Thus,  representing  the  arc  AP,  or  the  angle  AOP  by  6,  we 
have,  when  0  A  =  OP  =  1, 


;:fl 


'.  1' 


22 


PLANE  THIOONOMETRY. 


sintf  =^^  =  ^-^  =  PC, 


I 


m 


PC 

or 

PC 

1 

AT 

AT 

and  similarly  for  the  other  functions. 

Therefore,  in  a  circle  whose  radius  is  unity,  the  Trigono- 
metric functions  of  an  arc,  or  of  the  angle  at  the  centre  meas- 
ured by  that  arc,  may  be  defined  as  follows : 

The  sine  is  the  perpendicular  let  fall  from  one  extremity  of 
the  arc  upon  the  diameter  passing  through  the  other  extremity. 

The  cosine  is  the  distance  from  the  centre  of  the  circle  to  the 
foot  of  the  sine. 

TJie  tangent  is  the  line  which  touches  one  extremity  of  the 
arc  and  is  terminated  by  the  diameter  produced  passing  through 
the  other  extremity. 

The  secant  is  the  portion  of  the  diameter  produced  through 
one  extremity  of  the  arc  ivhich  is  intercepted  between  the  centre 
and  the  tangent  at  the  other  extremity. 

The  versed  sine  is  the  part  of  the  diameter  intercepted 
between  the  beginning  of  the  arc  and  the  foot  of  the  sine. 

Since  the  lines  PD  or  OC,  BT',  OT',  and  BD  are  respect- 
ively the  sine,  tangent,  secant,  and  versed  sine  of  the  arc 
BP,  which  (Art.  12)  is  the  complement  of  AP,  we  see  that 
the  cosine,  the  cotangent,  the  cosecant,  and  the  coversed  sine  of 
an  arc  are  respectively  the  sine,  the  tangent,  the  secant,  and  the 
versed  sine  of  its  complement. 


EXAMPLES. 

1.   Prove  tan  A  sin  A  +  cos  A  =  sec  A. 


2. 
3. 
4. 
6. 
6. 


(( 


<c 


(t 


cot  A  cos  A  +  sin  A  =  cosec  A. 

(tan  A  -  sin  A)^^-  (1  -  cos  A)^  =  (sej  A  -  ly. 

tan  A  -}-  cot  A  =  sec  A  cosec  A. 

(sinA+cos  A)-^-(sec  A  -f-  cosecA)=sinA  cos  A. 

(l  +  tanA)2-f  (l  +  cotA)2=(secA  +  coseoA)l 


POSITIVE  AND   NEGATIVE  LINES, 


28 


Trigono- 
re  meas- 

emity  of 
vtremity. 
de  to  the 

ty  of  the 
r  through 

through 
he  centre 

tercepted 
i  sine. 
respect- 
the  arc 
see  that 
i  sine  of 
,  and  the 


-ly. 

A  cos  A. 
secA)^. 


B 


M' 


O    M 


7.  Given  tanA  =  cot2  A;  find  A. 

8.  "      sin  A  =c()8.'}A;  liiid  A. 

9.  "      sin  A  =  cos  (45°-  ^  A)  ;  iind  A. 

10.  "      tanA  =  cotGA;  lind  A. 

11.  "      cot  A  =  tan  (45°+  A) ;  find  A. 

17.  Positive  and  Negative  Lines.  —  Let  A  A'  and  BB'  be 

two  periHMidicular  riglit  lines  intersecting  at  the  point  U. 
Then  the  position  of  any  point  in 
the  line  AA'or  1>B'  will  be  deter- 
mined if  we  know  the  distance  of 
the  point  from  O,  and  if  we  know 
also  upon  which  side  of  O  the  A— 
point  lies.  It  is  therefore  con- 
venient to  employ  the  algebraic 
signs  +  and  — ,  so  that  if  dis- 
tances measured  along  the  fixed 
line  OA  or  OB  from  O  in  one  direction  be  considered 
positive,  distances  measured  along  OA'  or  OB'  in  the  oppo- 
site direction  from  0  will  be  considered  negative. 

This  convention,  as  it  is  called,  is  extended  to  lines  parallel 
to  A  A'  and  BB';  and  it  is  customary  to  consider  distances 
measured  from  BB' towards  the  right  and  from  A  A'  upivards 
as  positive,  and  consequently  distances  measured  from  BB' 
towards  the  left  and  from  AA'  doivmvards  as  negative. 

18.  Trigonometric  Functions  of  Angles  of  Any  Magni- 
tude. —  In  the  definitions  of  the  trigonometric  functions 
given  in  Aru.  13  we  considered  only  acute  angles,  i.e.,  angles 
in  the  first  quadrant  (Art.  5),  since  the  angle  was  assumed 
to  be  one  of  the  acute  angles  of  a  right  triangle.  We  shall 
now  show  that  these  definitions  apply  to  angles  of  any  mag- 
nitude, and  that  the  functions  vary  in  sign  according  to  the 
quadrant  in  which  the  angle  happens  to  be. 


I 


24 


PLANE  TRIGONOMETRY. 


Let  AOP  be  an  angle  of  any  mag- 
nitude formed  by  OP  revolving  from 
an  initial  position  OA.  Draw  PM  ± 
to  AA'.  Consider  OP  as  always 
positive.  Let  the  angle  AOP  be 
denoted  by  A  ;  then  whatever  be  the 
magnitud'!  of  the  angle  A,  the  defini- 
tions of  the  trigonometric  functions 
are 


sin  A 


sec  A 


MP  .      OM 

,  cosA= , 

OP'  OP ' 


tan  A     = 


OP 
OM' 


cotA  = 


OM 
MP' 


cosecA  = 


I.  When  A  lies  in  the  1st  quadrant, 
MP  is  positive  because  measured  from  M 
iqnvards,  <^^M  is  positive  because  measured 
from  O  towards  the  right  (Art.  17),  and  OP 
is  positive. 

Hence  in  the  first  quadrant  all  the  func- 
tions are  positive. 

II.  When  A  lies  in  the  2d  quadrant,  as 
the  oh;  use  angle  AOP,  MP  is  positive 
because  meas"red  from  M  u^nvards,  OM  is 
negative  becf  ase  measured  from  0  towards 
the  left  (Art.  17),  and  OP  is  positive. 

Hence  in  the  second  quadrant 

sin  A  =  — —  is  positive  ; 


cos  A 


OM 


--— -  IS  negative : 
OP  ^         ' 


■        .      MP  . 

tsin  A  =^~:  IS  negative ; 
OM  •'         ' 

and  therefore  sec  A  and  cot  A  are  negative,  and  cosec  A  is 
positive  (Art.  13). 


FUNCTIONS  OF  ANGLFS. 


25 


III.  When  A  lies  in  the  3J  quadrant, 
as  the  reflex  angle  AOP,  MP  is  negative 
because  measured  from  M  downwards,  OM 
is  negative,  and  OF  is  jjositive. 

Hence  in  the  third  quadrant  the  sine, 
cosine,  secant,  and  cosecant,  are  negative, 
but  the  tangent  and  cotangent  are  positive. 

IV.  When  A  lies  in  the  4th  quadrant, 
as  the  reflex  angle  AOP,  MP  is  negative, 
OM  is  positive,  and  OP  is  positive. 

Hence  in  the  fourth  quadrant  the  sine, 
tangent,  cotangent,  and  cosecant  are  negative, 
but  the  cosine  and  secant  are  positive. 

The  signs  of  the  different  functions  are  shown  in  the 
annexed  table. 


Quadrant. 

I. 

II. 

• 
III. 

IV. 

Sin  and  cosec 

+ 

+ 

— 

Cos  and  sec- 

+ 

— 

— 

+ 

Tan  and  cot 

— 

+ 

— 

NOTB.  —  It  is  apparout  from  tliis  table  tbat  tbe  8ij;nB  of  all  the  functions  in  any 
quadrant  are  known  wlien  those  of  the  sine  am!  cosine  are  known.  The  tangent  and 
cotangent  are  +  or  — ,  according  as  the  sine  and  cosine  have  like  or  different  signs. 

19.  Changes  in  the  Value  of  the  Sine  as  the  Angle  in- 
creases from  0°  to  380°.  —  Let  A  de- 
note the  aiigle  AOP  described  by  the 
revolution  of  OP  from  its  initial  posi- 
tion OA  through  3G0°.  Then,  PM 
being  drawn  perpendicular  to  AA', 

sin  A  =  MI!, 
OP 

whatever   be  the  magnitude  of  the 

angle  A. 


26 


PL  A  NE  TRIGONOMETR  Y. 


\\ 


I    • 


When  the  angle  A  is  0°,  P  coincides  with  A,  and  MP  is 
zero;  therefore  sin 0°  =  0. 

As  A  increases  from  0°  to  90°,  MP  increases  from  zero  to 
OB  or  OP,  and  is  positive;  therefore  sin  90°=  1. 

Hence  in  the  1st  quadrant  sin  A  is  positive,  and  increases 
from  0  to  1. 

As  A  increases  from  90°  to  180°,  MP  decreases  from  OP 
to  zero,  and  is  p)ositive;  therefore  sin  180°  =  0. 

Hence  in  the  2d  quadrant  sin  A  is  positive,  and  decreases 
from  1  to  0.  ,  :;       \<  ■■ 

As  A  increases  from  180°  to  270°,  MP  increases  from 
zero  to  OP,  and  is  negative;  therefore  sin 270°  =  —  1. 

Hence  in  the  3d  quadrant  sin  A  is  negative,  and  decreases 
algebraically  from  0  to  —  1 . 

As  A  increases  from  270°  to  3G0°,  MP  decreases  from  OP 
to  zero,  and  is  nega4ive  ;  therefore  sin  3G0°  =  0. 

Hence  in  the  4tli  quadrant  sin  A  is  negative,  and  increases 
algebraically  from  —  1  to  0. 


f 


20.  Changes  in  the  Cosine  as  the  Angle  increases  from  0° 


to  360°.  —  In  the  figure  of  Art.  19 


cosA  = 


OM 

op' 


When  the  angle  A  is  0°,  P  coincides  with  A,  and 
OM  =  OP  ;  therefore  cosO°=  1. 

As  A  increases  from  0°  to  90°,  OM  decreases  from  OP  to 
zero  and  is  j^ositive;  therefore  cos  90°=  0. 

Hence  in  the  1st  quadrant  cos  A  is  positive,  and  decreases 
from  1  to  0. 

As  A  increases  from  90°  to  180°,  OM  increases^from  zero 
to  OP,  and  is  negative;  therefore  cos  180°  =—  1.'  ' 

Hence  in  the  2d  qunlrant  cos  A  is  negative,  and  decreases 
algebraically  from  0  to  —  1. 

As  A  increases  from  180°  to  270°,  OM  decreases  from  OP 
to  zero,  and  is  negative  ;  therefore  cos  270°  =  0. 


t 


CHANGES  IN   THE  TANGENT. 


27 


Hence  in  the  3d  quadrant  cos  A  is  negative,  and  increases 
algebraically  from  —1  to  0. 

As  A  increases  from  270°  to  360°,  OM  increases  from  zero 
to  OP,  and  is  positive;  therefore  cos 360°=  1. 

Hence  in  the  4th  quadrant  cos  A  is  positive,  and  increases 
from  0  to  1. 

21.  Changes  in  the  Tangent  as  the  Angle  increases 
from  0°   to   360°.  —  In  the  figure  of  Art.  19 

tan  A  =  — -• 
OM 

When  A  is  0°,  MP  is  zero,  and  OM  =  OP;  therefore  tan 
0°=0. 

As  A  increases  from  0°  to  90°,  MP  increases  from  zero  to 
OP,  and  OM  decreases  from  OP  to  zero,  so  that  on  both 
accounts  tan  A  increases  numerically;  therefore  tan90°  =  QO. 

Hence  in  the  1st  quadrant  tan  A  is  positive,  and  increases 
from  0  to  cc. 

As  A  increases  from  90°  to  180°,  MP  decreases  from  OP 
to  zero,  and  is  positive,  OM  becomes  negative  and  decreases 
algebraically  from  zero  to  —  1 ;  therefore  tan  180°=  0. 

Hence  in  the  2d  quadrant  tan  A  is  negative,  and  increases 
algebraically  from  -co  to  0. 

When  A  passes  into  the  2d  quadrant,  and  is  only  just  greater  than 
90*^,  tan  A  changes  from  +co  to  —  oo. 

As  A  increases  from  180°  to  270°,  MP  increases  from  zero 
to  OP,  and  is  negative,  OM  decreases  from  OP  to  zero,  and 
is  negative;  therefore  tan 270°=  oo. 

Hence  in  Liie  3d  quadrant  tan  A  is  positive,  and  increases 
from  0  to  00. 

As  A  increases  from  270°  to  360°,  MP  decreases  from  OP 
to  zero,  and  is  negative,  OM  increases  from  zero  to  OP,  and 
is /)osrt*'»;e;  therefore  tan  360°=  0. 

Hence  in  the  4th  quadrant  tan  A  is  negative,  and  increases 
algebraically  from  —  oo  to  0. 


28 


PLANE  TRIGONOMETRY. 


w 


The  student  is  recommended  to  trace  in  a  manner  similar 
to  the  above  the  changes  in  the  other  functions,  i.e.,  the 
cotangent,  secant,  and  cosecant,  and  to  see  that  his  results 
agree     ith  those  given  in  the  following  table. 


22.    Table    giving    the    Changes    of    the   Trigonometric 
Functions  in  the  Four  Q,uadrants. 


Quadrant. 

I. 

II. 

III. 

IV. 

sill  varies  from 

+ 
Oto  1 

+ 

1  too 

Oto  -  1 

-1  too 

cos      "        " 

+ 
1  toO 

0  to  -  1 

-  1  too 

+ 
Oto  1 

tan      "        " 

+ 
0  to  CO 

-00  toO 

0  to  00 

-00  toO 

cot      "        " 

+ 
00  toO 

oto  00 

4- 
00  to  0 

0  to  —  00 

sec      "        " 

+ 

1  to  00 

—  00  to  —  1 

—  1  to   —  00 

00  to  1 

cosec  "        " 

+ 

00  to  1 

+ 
1  to  00 

—  00  to  —  1 

—  1  to   —  00 

vers    "        " 

+ 

oto  1 

4 
1  to  2 

+ 

2tol 

1  toO 

'i 


Note  1.  —  The  cosecant,  secant,  and  cotangent  of  an  angle  A  have  the  same  sign 
as  the  sine,  cosiiie,  and  tangent  of  A  respectively. 

The  sine  and  cosine  vary  from  1  to  — 1,  passing  through  the  value  0.  They  are 
never  greater  than  tinity. 

The  secant  and  cosecant  vary  from  1  te  — 1,  passing  through  the  value  oo.  They 
are  never  mimerically  less  than  unity. 

The  tangent  and  cotungent  are  unlimited  In  value.    They  have  all  values  from  —  w 

to  +00. 

The  versed  sine  and  coversed  sine  vary  from  0  to  2,  and  are  always  positive. 

The  trigonometric  functions  change  sign  in  passing  through  the  values  0  and  », 
and  through  no  other  values. 

In  the  1st  quadrant  the  functions  increase,  and  the  coJ\tnctiona  decrease. 

Note  2.  —  From  the  results  given  in  the  ahove  table,  it  will  be  seen  that,  if  the 
value  of  a  trigonometric  function  be  Riven,  we  cannot  fix  on  one  angle  to  which  it 
belongs  exclusively. 

Thus,  if  the  given  value  of  sin  A  be  '  .ve  know  since  sin  A  passes  through  all 
values  from  0  to  1  as  A  increases  from  i    to  90°,  that  one  value  of  A  lies  between  0° 


RELATIONS  BETWEEN  FUNCTIONS. 


29 


and  90°.    But  since  we  also  know  that  the  value  of  sin  A  paaaes  through  all  values 
between  1  and  0  aa  A  increases  from  90"^  to  180°,  it  is  evident  that  there  is  another  ♦ 

value  of  A  between  90°  and  180°  for  which  sin  A  =  J, 


23.  Relations  between  the  Trigonometric  Functions  of 
the  Same  Angle.  —  Let  the  radius  start 
from  the  initial  position  OA,  and  revolve  p^ 
in  either  direction,  to  the  position  OP. 
Let  6  denote  the  angle  traced  out,  and 
let  the  lengths  of  the  sides  PM,  MO, 
OP  be  denoted  by  the  letters  a,  b,  c* 

The  following  relations   are   evident 
from  the  definitions  (Art.  13)  : 

cosec  0  =  — — ,  sec  6  = ,  cot  $  = 


tan  6  = 


sin^ 

sin  0 
cos^ 


cos^ 


ta,nO 


For 


tan^: 


a 
b 


a 

c 
l 
c 


sin  0 

» 

cos  6 


IL    sin2^  +  cos2^  =  L 
For  sin'^  +  cos^e 

IIL  sec''^=l  +  tan2  0 


^  +  ^  =  ^'  +  &' 


For  sec^^=^^  =  *-^  =  l  +  ^^=l  +  tan»^. 


IV.  cosec''^  =  l  +  cot2  6. 


For  cosec*^=— = 


a^  +  ft' 


a' 


a" 


=  l+-T=l-fC0t2ft 


a* 


Formulae  I.,  IL,  IIL,  IV.  are  very  important,  and  must  be 
remembered. 

*  a,b,c  are  numbers,  being  the  number  of  times  the  lengths  of  the  sides  contain 
some  chosen  uait  of  length. 


i   ll 

If! 
I    li 


80 


PLANE  TRIGONOMETRY. 


24.   Use  of  the  Preceding  Formulee. 

1.    To  express  all  the  other  functions  in  terms  of  the  sine. 

Since      sin^^ +  cos^^  =  1,     .-.  cos6  =  ±Vl 

sin^       .         sin^ 


sin^d. 


tan  6  = 

cot^  = 

sec^: 
cosec  6  ■■ 


cos  6 

1 

tand 

1 

cos  6 

1 
sind 


=  ± 
=  ± 


VI- sin- ^ 

VI  -  sin2^ 
sin^ 

1 

Vl-sin^^ 


II.    To  express  all  the  other  functions  in  terms  of  the  tan- 
gent. 

sin  ^ 


Since     tan  B  = 


cos 


^ 


sine  =  tand  cos  6  =  ^^  =  ± 


tan^ 


cose  = 
^  cote  = 
cosec  e  = 


sec^ 

1 
tan^ 

1 

sin  6 


=  ± 


seed 
1 


Vl  +  tan^d 


Vl  +  tan^d 
seed  =  ±  Vi  +"tai?^. 


=  ± 


Vl+tan^d 
tan  $ 


Similarly,  any  one  of  the  functions  of  an  angle  may  be 
expressed  in  terms  of  any  other  function  of  that  angle. 
The  sign  of  the  radical  will  in  all  cases  depend  upon  the 
quadrant  in  which  the  angle  0  lies. 

25.  Graphic  Method  of  finding  All  the  Functions  in  Terms 
of  One  of  them.  ^B 

To  express  all  the  other  functions  in 
tei'ms  of  the  cosecant. 

Construct  a  right  triangle  ABC,  hav- 
ing the  side  BC  =  1.     Then  a       Vcoaeca-l 


RELATIONS  BETWEEN  FUNCTIONS. 


31 


.       AB      AB       .^ 

cosec  A  =  - — -  =    -~  =  AB. 


Now 


•.  AC  =  ±  Vcosec''*  A  —  1. 
smA  =  — — -  = 


AB      cosec  a' 


A       AC       ,  Vcosec^A  — 1 
cos  A  =  —  =  ±  ~ , 

AB  cosec  A 

tanA  =  ?^^  ^ 

AC 


Vcosec'^A  — 1 

and  similarly  the  other  functions  may  be  expressed  in  terms 
of  cosec  A. 

26.  To  find  the  Trigonometric  Functions  of  46".  —  Let 

ABC  be  an  isosceles  right  triangle  in  which  g 

CA  =  CB. 

Then        CAB  =  CBA  =  45°. 

Let  AC  =  m  =  CB. 

Then 


:2         T-7^2 


AB'  =  AC  4-  CB'  =  m"  +  m^  =  2  m\ 
.-.  AB  =  mV2. 


sin45°  =  5C  = 
AB 

cos45°  =  4^  = 


m 


m  V2      V2 
m  1 


AB     .„iV2      V2 
tan45°=?5  =-  =  L     cot45°=l. 

AC        771 

sec45°=V2.     cosec45°=V2. 


27.  To  find  the  Trigonometric  Func- 
tions of  60°  and  30°. —Let  AB  be  an 

equilateral  triangle.  Draw  AD  perpen- 
dicular to  BC.  Then  AD  bisects  the 
angle  BAC  and  the  side  BC.  Therefore 
BAD  =  30°,  and  ABD  =  00°. 


B      fn 


1     • 

■ 

< 

j 
1 

'^ 

1 

4[ 

i' 

Ill: 


32 


PLANE  TRIGONOMETRY, 


Let 

BA  =  2m.     .-.  BD  =  m. 

Then 

AD  =  V4  mi^  —  m'^  =  »?i  V3. 

•  • 

3in60°-^^-^;/^-W3.    .-.  cosec60°      ^ 
AB       2  m                                          V3 

cos60"=.^^  =  ^.       .-.  sec60°  =  2. 
BA      2 

tan60°-^|^-^^^-V3.     .-.  cot60°_    1. 
BD         m                                        yg 

vers«0°=  1  -  cos  60°=  1-1  =  1.      ^  '       V  "     ' 

Also 

sin30°  =  ^  =  ^  =  ^.     .-.  cosec30°=2.  r 
AB     2m     2 

cos30°=-— —  =  — ^ —  —  iv3.    .-.  sec30°=       • 
AB       2  m                                        ^3 

tan30°— ^"^—     '"*'     =    ^  .      •    oc^'\(\°^-k/^ 

DA       ™  a/:^       a/'? 

;  EXAMPLES. 

3 
1.    Given    sin  ^  =  - ;   find    the    other 

trigonometric  functions. 

Let  BAG  be  the  angle,  and  BC  be 
perpendicular  to  AC.      Represent  BC 
by  3,  AB  by  5,  and  consequently  AC  ^ 
by  V26^^  =  4. 

AC     4 


Then 


cos^ 


AB 

5' 

*-^=Ic  = 

3 

.e.  =  f-«. 

5 

=  4' 

cosec0  =  ||  = 

5 
"3' 

REDUCTION   OF  FUNCTIONS. 


33 


3  3    5 

2.    Given  sin 6  =  -:  find  tan 6  and  cosec 6.      Aiis.   -)    -• 
6  4    3 


3.  Given  cos  9  =  --,  find  sin  6  and  cot  0. 

o 

4.  Given  sec^  =  4;  find  cot  ^  and  sin  ^. 

5.  Given  tan  6  =  V3 ;  find  sin  (9  and  cos  0. 

'"'^'  12 

6.  Given  sin  ^  =  t^  ;  find  cos^. 

7.  Given  cosec  ^  =  5 ;  find  sec  6  and  tan  B. 


'         2V2 

1      Vlii 

VTs'    4 

W3,| 

_5 
13* 


5 


2V()  2VG 
40  J)^ 
4l'    41' 

V5    3 


3 


8.  Given  sec  0  =  — ;  find  sin  6  and  cot  0. 

2 

9.  Given  cot  6  =  — ;  find  sin  6  and  sec  0. 

V5 

3 

10.    Given  sin  0  =  -  ;  find  cos  6,  tan  6,  and  cot  ^. 

,       ^  V7    3V7     V7 

.:;,-\,.,..  .-^  X'   "T"'    T"' 

6 


11.    Given  sin  0  =  - ;  find  tan  6. 

c 


12.    Given  sin0  = 


2  mil 


;  find  tan^. 


2m?i 
m-—  11- 


28.  Eeduction  of  Trigonometric  Functions  to  the  Ibt 
Quadrant.  — All  mathematical  tables  give  the  trigonometric 
functions  of  angles  between  0°  and  90°  only,  but  in  practice 
we  constantly  have  to  deal  with  angles  greater  than  90°. 
The  object  of  the  following  six  Articles  is  to  show  that  the 
trigonometric  functions  of  any  angle,  positive  or  negative, 
can  be  expressed  in  terms  of  the  trigonometric  functions  of 
an  angle  less  than  90°,  so  that,  if  a  given  angle  is  greater 
than  90°,  we  can  find  an  angle  in  the  1st  quadrant  whose 
trigonometric  function  has  the  same  absolute  value. 


V       -« 


84 


P,LANE  TRiaONOMETliy. 


1  t 


I'll 
iii 

\     'a 

■r,i   , 


29.  Functions  of  Complemental  Angles.  —  Let  AA',  HB' 

be  two  diaiuuters  of  a  circle  at  right  angles,  and  let  OP  and 
OP'  be  the  positions  of  the  radins  for  b 

any  angle  AOP  =  A,  and  its  comple- 
ment A0P'=1)()°- A  (Art.  12). 

Draw  PM  and  P'M'  at  right  angles  J 
to  OA. 

Angle  OP'M'=  B0P'=  AOP  =  A. 

Also  0P=  OP'. 

Hence  the  triangles  0PM  and  OP'M'  are  equal  in  all 

respects. 

P'M'  ^  OM 

~  op" 


P'M'=  OM. 


01 


jf 


Also, 


sin  (90°  -  A)  =  cos  AOP  =  cos  A. 

OM'  ^  PM 
~0P* 


OM'  =  PM. 


01 


Jf 


cos  (90°  -  A)  =  sin  AOl'  =  sin  A. 


Similarly,     tan  (90°-  A)  =  tan  AOP'=  :!-^  =  1±!^  =  cot  A. 

The  other  relations  are  obtained  by  inverting  the  above. 

30.  Functions  of  Supplemental  An- 
^  gles,  —  Let  OP  and  OP'  be  the  positions 
of  the  radius  for  any  angle  AOP  =  A,    p 
and  its   supplement  AOP'=180°-A    ,/ 
(Art.  12). 

Since  OP  =  OP',  and  POA  =  P'OA', 
the  triangles  POM  and  P'OM'  are 
geometrically  equal. 

P'M'      PM 
.-.  8in(180°-A)=sinAOP'=-i-^=^  =  sinA, 


OP 


cos  (180° -  A)  =  cos  AOP'= 


OM' 
OP' 


OM 


OP 


=  —  cos  A, 


REDUCTION   OF  FUNCTIONS. 


35 


tan  (180°-  A)  =  tan  AOr'=  ^-^  =  ~  =  -  tan  A. 

OM'      OM 

Similarly  the  other  relations  may  bo  obtained. 


31.  To  prove  8in(00°+A)  =  co8A,  cos  (00°  +  A)  =  — sin  A, 
and  tan  (90°  +  A)  =  -  cot  A. 

Let  OP  and  OP'  be  the  positions 
of  the  radius  for  any  angle  AOP  =  A, 
and  AOP'=90°+A. 

Since  OP  =  OP',  and  AOP  =  P'OB 
=  OP'M',  the  triangles  POM  and 
P'OM'  are  equal  in  all  respects. 

.-.  sin  (90°+  A)  =  sin  AOP'==  -^  =  ^^  =  cos  A, 

cos  (90°+  A)  =  cos  AOP'=  ^'  =  — —  =  -  sin  A, 
^  ^  OP'         01' 

.     tan(90°+A)  =  tanA0P'=-^'=-^^^  =  -cotA. 
^  ^  OM'      -PM 


32.  To  prove  sin  ( 180°  +  A)  =  - sin  A, 
cos  (180°+  A)  =-  -  cos  A,  and  tan  (180° 
+  A)  =  tanA. 

Let  the  angle  AOP  =  A ;  then  the 
angle  AOP',  measured  in  the  positive 
direction,  =  (180°+ A).  p 

The  triangles  POM  and  P'OM'  are 
equal. 


.-.  sin  (180°+ A)  =  sin  AOP' = 


P'M' 
OP' 

OM' 


cos  (180°+  A)  =cos  AOP'  =  ^^  = 
tan  (180°+  A)=tanAOP'  = 


OP' 
P'M' 


OM'      -  OM 


36 


PL  A  NE   TRIG  ONOMETR  T. 


l:,l 


>      1 


33.    To    prove    sin  ( —  A)  =  —  sin  A,    cos  ( —  A)  =  cos  A, 

tan  ( —  A)  =  —  tan  A.  B 

Let  OP  and  01"  be  the  positions  of 
the  radius  for  any  tuiual  angh>s  AOP 
and  AOP'  measured  from  the  initial 
line  AO  in  opposite  directions.  Then 
if  the  angle  AOP  be  denoted  by  A,  the 
numerically  equal  angle  AOP'  will  be 
denoted  by  —  A  (Art.  4). 

The  triangles  POM  and  P'OM  are  geometrically  equal. 


.-.  sin(-A)  =  .sinAOP'  = 


PMVt 
OP'^ 


PM 


OP 


=  —  sin  A, 


cos  (  — A)  =  cos  AOP' =  —~-  =  — —  =  cosA, 

tan  ( -  A)  =  tan  AOP'  =  ^^^  =  ~-^^  =  -  tan  A. 
^        ^  OM        OM 


34.  To   prove   8in(270°  + A)  =  sin(270°- A)  =  -co8A, 
and  cos  (270°  + A)  =  - cos  (270°- A)  g 

=  8inA. 

Let  the  angle  AOP  =  A ;  then  the       / 
angles  AOQ  and  AOR,  measured  in 
the   positive    direction,    =(270°— A)  ^\ 
and  (2^0°+ A)  respectively. 

The  triangles  POM,  QON,  and  EOL 
are  geometrically  equal.  ^ 


EL  =  QX     =  OM. 


^-  =  QN^-J)M 
OE     OQ        OP 


Also, 


.-.  sin  (270°+  A)  =  sin  (270°-  A)  =  -  cos  A. 
;  OL      -ON      PM 


OE 

.-.  cos  (270°  + A): 


OQ        OP 

-  cos  (270°- A)  =  sin  A. 


h 


VALUES  OF  THE  FUNCTIONS. 


37 


35.  Table  giving  the  Values  of  the  Functions  of  Any  Angle 
in  Terms  of  the  Functions  of  an  Angle  less  than  90".  —  The 

foregoing  results,  and  other  similar  ones,  which  may  be 
proved  in  the  same  manner,  are  here  collected  for  reference. 

Quadrant  II. 
sm(180°-A)=      sinA.  sin(90°+A)=     cosA.' 


cos(180°- A)  =  — cosA. 
tan  (180°- A)  =  - tan  A. 
cot  (180°- A)  =  - cot  A. 
sec  (180°- A)  =  - sec  A. 
cosec  (180°—  A)  =  cosec  A. 


cos(90°+ A)  =  -sinA. 
tan(00°+ A)  =  -cotA. 
cot(9()°-f  A)=.-tanA. 
sec  (1)0°  +  A )  =  —  cosec  A. 
cosec  (90°  -f  A )  =      sec  A. 


Quadrant  III. 


sin  ( 180°+ A)  = 
cos  (180°+ A)  = 
tan  (180°+ A)  = 
cot  (180°+ A)  = 


sin  A. 
cos  A. 
tan  A. 
cot  A. 
sec  A. 


sin  (270°- A)  = 
cos  (270°- A)  = 
tan  (270°- A)  = 
cot  (270°- A)  = 
sec  (270°- A)  = 


sec  (180°+ A)  = 
cosec  (180°+  A)  =  -  cosec  A.   cosec  (270°-  A)  = 


—  cos  A. 

—  sin  A. 
cot  A. 
tan  A. 

—  cosec  A. 

—  sec  A. 


Quadrant  IV. 
A)  =  -sinA.  sin(270°+ A)  =  -cosA. 

cos  (270°+ A)  =  sin  A. 
tan(270°+ A)  =  -cotA. 
cot  (270°+ A)  =  - tan  A. 
sec  (270°+ A)  =      cosec  A. 


sin  (SCO' 

cos  (360°- A)  =  cos  A. 
tan(360°- A)  =  -tanA. 
cot(360°- A)=:— cotA. 
sec  (360°- A)  =  sec  A. 


cosec  (360°—  A)  =  —  cosec  A.  cosec  (270°+  A)  =  —  sec  A. 

Note.  —  These  relations*  may  be  remembered  by  noting  the  followinp:  rules  : 

When  A  \»  associated  with  an  ei'en  multiple  of  90°,  any  function  of  the  angle  is 
numerically  equal  to  the  same,  function  of  A. 

When  A  is  associated  with  an  odd  multiple  of  90°,  any  function  of  the  angle  is 
uumericaliy  equal  to  the  corresponding  cofunction  of  the  angle  A. 

The  sign  to  be  prefixed  will  depend  upon  the  quadrant  to  which  the  angle  belongs 
(Art.  5),  regarding  A  as  an  acute  angle. 


*  Although  these  relations  have  been  pro%'ed  only  in  case  of  A,  an  acute  angle, 
they  are  true  whatever  A  may  be. 


I) 
i 


88 


PLANE  TRIGONOMETRY. 


;i 

■ 

ij 

il 

■ 

hi 


'  !l! 

i  ■  li 
I  'i 


ThuB,  cos  (270°  -A)  =  — sinA;  the  angle  270°  — A  being  in  the  3d  quadrant,  and 
its  cosine  uegiUivc  in  consequence. 
For  an  angle  in  the 

i^^irst  qiittdrant  all  the  functions  are  positive. 

Second  quadrant  all  are  negative  except  the  sine  and  cosecant. 

Third  quadrant  all  are  negative  except  the  tangent  and  cotangent.  J 

Fourth  quadrant  all  are  negative  except  the  cosine  and  secant. 

36.  Periodicity  of    the  Trigonometric  Functions.  —  Let 

AOP  be  an  angle  of  any  magnitude,  as  in  the  figure  of 
Art.  18 ;  then  if  OP  revolve  in  the  positive  or  the  negative 
direction  through  an  angle  of  360°,  it  will  return  to  the 
position  from  which  it  started.  Hence  it  is  clear  from  the 
definitions  that  the  trigonometric  functions  remain  un- 
changed when  the  angle  is  increased  or  diminished  by  360°, 
or  an}'  multiple  of  360°.  Thus  the  functions  of  the  angle 
400°  are  the  same  both  in  numerical  value  and  in  algebraic 
sign  as  the  functions  of  the  angle  of  400°—  360°,  i.e.,  of  the 
angle  of  40°.  Also  the  functions  of  360°+ A  are  the  same 
in  numerical  value  and  in  sign  as  those  of  A. 

In  general,  if  w  denote  any  integer,  either  positive  or 
negative,  the  functions  of  n  x  360°+  A  are  the  same  as  those 
of  A. 

Thus  the  functions  of  1470°=  the  functions  of  30°. 

If  0  denotes  any  angle  in  circular  measure,  the  functions 
of  (27i7r  +  0)  are  the  same  as  those  of  0.     Thus 

sin  (2w7r  +  d)  =  s'mO,  cos  (2 wtt  -\-6)  =  cos  0,  etc. 

By  this  proposition  we  can  reduce  an  angle  of  any  magni- 
tude to  an  angle  less  than  360°  without  changing  the  values 
of  the  functions.  It  is  therefore  unnecessary  to  consider 
the  functions  of  angles  greater  than  360° ;  the  formulae 
already  established  are  true  for  anyles  of  any  magnitude 
whatever. 

EXAMPLES. 

Express  sin  700°  in  terms  of  the  functions  of  an  acute 
angle. 

sin 700°=  sin  (360°+  340°)  =  sin 340°=  sin  (180°+  160°) 
=  sinl60°=~sin2()°. 


l«Mi 


•«»« 


EXAMPLES, 


39 


Express  the  following  functions  in  terms  of  the  functions 
of  acute  angles : 

1.  sin 204°,  sin 510°.  Aiis.   -sin 24°,  sin 30°. 

2.  cos  (-800°),  cos  359°.  cos  80°,  cosl°. 
G.   tan 500°,  tan 300°.  -tan 40°,   -cot 30°. 

Find  the  value  of  the  sine,  cosine,  and  tangent  of  the 
following  angles : 


4. 

150°. 

5. 

-  240°. 

C. 

330°. 

•  7. 

225°. 

Ans. 


iV3, 


iV3, 

1 

"2' 


1 

V3' 

-V3. 


2  V3 

._JL     _1_     1 

V2         V2 
Find  the  values  of  the  following  functions  : 

8.    sin 810°,  sin (-240°),  cos  210°.     ^><6-.  1,  |V3,  -^VS. 


.  9.    tan  (-120°),  cot  420°,  cot  510° 

10.  sin  930°,  tanG420°. 

11.  cot  1035°,  cosec570°. 


V3,    -^,    1. 
V3 
__1     J_ 

2'    V3* 
-1,  -2. 


37.  Angles  corresponding  to  Given  Functions.  —  When  an 
angle  is  given,  we  can  hud  its  trigonometric  functions,  as  in 
Arts.  26  and  27 ;  and  to  each  value  of  the  angle  there  is 
but  one  value  of  each  of  the  functions.  But  in  the  converse 
proposition  —  being  given  the  value  of  the  trigonometric 
functions,  to  find  the  corresponding  angles  —  we  have  seen 
(Art.  36)  that  there  are  many  angles  of  different  magnitude 
which  have  the  same  functions. 

If  two  such  angles  are  in  the  .same  quadrant,  they  are 
represented  geometrically  by  the  same  position  of  OP,  so 
that  they  differ  by  some  multiple  of  four  right  angles. 


i 


40 


PLANE  TRIGONOMEJRY. 


^il 


ri^'i 


If  we  are  given  the  value  of  the  sine  of  an  angle,  it  is 
important  to  be  able  to  find  all  the  angles  which  have  that  ' 
value  for  their  sine.  ,        , . 

38.  General  Expression  for  All  Angles  which  have  a 
Given  Sine  a.  —  Let  O  be  the  centre  of  the  unit  circle. 
Draw  the  diameters  AA',  BB',  at  right 
angles. 

From  0  draw  on  OB  a  line  ON,  so 
that  its  measure  is  a. 

Through  N  draw  PP'  parallel  to 
AA'.  Join  OP,  OP',  and  draw  PM, 
P'M',  perpendicular  to  AA'. 

Then  since  MP  =  M'P' =  ON  =  a, 
the  sine  of  AOP  is  equal  to  the  sine  of  AOP'. 

Hence  the  angles  AOP  and  AOP'  are  supplemental  (Art. 
30),  and  if  AOP  be  denoted  by  a,  AOP'  will  =  tt  -  «. 

Now  it  is  clear  from  the  ligure  that  the  only  2>ositive 
angles  which  have  the  sine  equal  to  a  are  «  and  tt  —  a,  and 
the  angles  formed  by  adding  any  multiple  of  four  right 
angles  to  a  and  tt  —  «.  Hence,  if  6  be  the  general  value  of 
the  required  angle,  we  have  .  ;- 

6  =  2mr-^a,  or  6  =  2mr -\-7r  —  a,     ....     (1) 

where  n  is  zero  or  any  positive  integer. 

Also  the  only  negative  angles  which  have  the  sine  equal 
to  a  are  —  (7r  +  «)j  and  —  {2v  —  n),  and  the  angles  formed 
by  adding  to  these  any  multiple  of  four  right  angles  taken 
negatively  ;  that  is,  we  have 

e=27i7r-(Tr  +  a),   6  =  2mr  -  {2ir  -  a),     .      .      (2) 

where  n  is  zero  or  any  negative  integer. 

Now  the  angles  in  (1)  and  (2)  may  be  arranged  thus: 

2rnr  +  a,   (2n  +  l)7r  — «,  (2n  — l)7r  — «,   {2n—2)ir  +  a, 

all  of  which,  and  no  others,  are  included  in  the  formula 

^  =  «7r +  (-!)"«, (3) 


GENERAL  EXPRESSION  FOR  ANGLES. 


41 


where  n  is  zero,  or  any  positive  or  negative  integer.  There- 
fore (3)  is  the  general  expression  for  all  angles  which  have 
a  given  sine. 

Note.  —  The  same  formula  dctertnineB  all  the  angles  wbicb  have  the  same 
cosacant  as  a. 

39.  An  Expression  for  All  Angles  with  a  Given  Cosine  a.  — 

Let  0  be  the  centre  of  the  unit  circle.  g 

Draw  AA',  BB',  at  right  angles. 

From  O  draw  OM,  so  that  its  meas- 
ure is  a. 

Through    M   draw   PP'   parallel  to 
BB'.     Join  OP,  OP'. 

Then  since  OM  —  a,  the  cosine  of 
AOP  is  equal  to  the  cosine  of  AOP'. 

Hence,  if  AOP  =  «,  A()P'=-«. 

Now  it  is  clear  that  the  only  angles  which  have  the  cosine 
equal  to  a  are  a  and  —  a,  and  the  angles  which  differ  from 
either  by  a  multiple  of  four  right  angles. 

Hence  if  9  be  the  general  value  of  all  angles  whose  cosine 
is  a,  we  have 

6  =  2mr±a,  "-  '  '      ; 

where  n  is  zero,  or  any  positive  or  negative  integer. 

Note.  —  The  same  formula  determines  all  the  angles  which  have  the  same  secant 
or  the  same  versed  sine  as  a. 


40.  An  Expression  for  All  Angles  with  a  Oiven  Tangent  a. 

—  Let   O   be  the  centre  of   the  unit  .-p 

circle.  Draw  AT,  touching  the  circle 
at  A,  and  take  AT  so  that  its  measure 
is  a. 

Join   OT,   cutting  the   circle   at  PA'j- 
and  P'. 

Then  it  is  clear  from  the  figure  that 
the  only  angles  which  have  the  tan-       ^ 
gent  equal  to  a  are  a  and  tt  -f  a,  and  the  angles  which  differ 


42 


PLANE  TRIGONOMETRY. 


w 


;     I 


from  either  by  a  multiple  of  four  right  angles.     Hence  if  6 
be  the  general  value  of  the  required  angle,  we  have 


0  =  2  iiTT  +  «,  and  2  mr  +  tt  +  «. 


(1) 


Also,  the  only  negative  angles  which  have  the  tangent 
equal  to  a  are  —  (tt  —  «),  and  —  (27r  — «),  and  the  angles 
which  differ  from  either  by  a  multiple  of  four  right  angles 
taken  negatively  ;  that  is,  we  have 

^  =  27177 —  (tt —«),  and  2mr  —  {2tt  —  a),  .     .     (2) 

where  n  is  zero  or  any  negative  integer. 

Now  the  angles  in  (1)  and  (2)  may  be  arranged  thus : 

2n7r  +  «,    (2?l  +  l)7r4-«,    (2?J  —  l)7r4-«,    (2?J-2)7r  +  «, 

all  of  which,  and  no  others,  are  included  in  the  formula 

^  =  W7r  +  «, (3) 

where  n  is  zero,  or  any  positive  or  negative  integer.  There- 
fore (3)  is  the  general  expression  for  all  angles  which  have 
a  given  tangent. 

Note.  — The  same  formula  determines  all  the  angles  which  have  the  same  cotan- 
gent as  a. 

EXAMPLES. 

1,   Find  six  angles  between   —  4  right  angles  and   -f  8 
right  angles  which  satisfy  the  equation  sin  A  =  sin  18°. 
We  have  from  (3)  of  Art.  38, 

e  =  n7r +  {-!)"—,  or  A  =  «xl80°+(-l)»18°. 

Put  for  n  the  vahu's  —  2,  —  1,  0,  1,  2,  3,  successively, 
and  we  get  A  =  -  360°  +  18°,  -  180°  -  18°,  18°,  180°  -  18°, 
360°+ 18°,  640°- 18°; 


that  is, 


342°,  -198°,  18°,  162°,  378°,  522° 


NoTK.  —  The  student  should  draw  a  figure  in  the  above  example,  and  in  each 
example  of  this  Liud  which  he  workt). 


TRIGONOMETRIC  IDENTITIES. 


43 


2.   Find   the   four    smallest    angles   which    satisfy    the 

equations  (1)  sinA  =  -,   (2)  sinA  =  — ,   (3)  sinA=  ---' 
^  2  -y/2  ■" 


(4)  siuA  =  -- 


Ans.  (1)  30°,  150°,  -  210°,  -  330° 

(2)  45°,  135°,  -225°,  -315° 

(3)  00°,  120°,  -  240°,  -  300° 

(4)  -30°,  -150°,  210°,  330°. 


41.  Trigonometric  Identities. — A  trigonometric  identitj'^ 
is  an  expression  which  states  in  the  form  of  an  equation  a 
relation  which  is  true  for  all  values  of  the  angle  involved. 
Thus,  the  relations  of  Arts.  13  and  23,  and  all  others  that 
may  be  deduced  from  them  by  the  aid  of  the  ordinary 
formulae  of  Algebra,  are  universally  true,  and  are  therefore 
called  identities;  but  such  relations  as  sin ^  =  4,  cosO=i, 
are  not  identities. 

'  •     EXAMPLES, 

1.    Prove  that     sec  6  —  tan  6  •  sin  0  =  cos  6. 


Here        see  $  —  tan^  sin  6  = ' sin  d 

cos  6     cos  0 

^l-sin''g 
cos  6 

cos^d 


(Art.  24) 


(Art.  24) 


cosd 
=  cos  6. 
2.   Prove  that     cot  ^  —  sec  ^  cosec  ^  (1  —  2  sin- 6/)  =  tan  B. 
cot^  — seed  cosecd(l  —  2  sin^^) 
cos^         1         1 


sin^      cos^  sin  ^ 

cos'^g-l  +  2sin'-tf 
sin  ^  cos  61 


(l-2sin-^)  (Art.  24) 


44 


PLANE  TRIGONOMETHY, 


l! 


Ill 


l':i! 


''\   ^! 


cos^^j-(siii-  ^  +  cos-  ^1+2^^ 
sill  0  cos  ^ 


(Art.  24) 


sin^ 
sin  9  cos  ^ 


siii^ 
cos  0 


=  tan  ^. 


Note.  —  It  will  be  observed  that  in  Bolving  Uieee  examples  we  first  express  the 
other  fiinctioiiH  in  tcrnia  of  the  sine  and  coHine,  und  in  most  cases  the  beginner  will 
find  this  the  siinplest  course.  It  is  generally  advisable  to  begin  with  the  most  com- 
pliculed  side  and  work  towards  the  otlier. 

Prove  the  following  identities  : 

o.  cos  6  tan  ^  =  si n  ^. 

4.  cos^  =  sin^  cot0. 

5.  (tan^  +  cot^),sin^cos^  =  l. 

6.  (taii^  —  cot6)  sin^cos^  =  sin'*^  —  cos^^. 

7.  siii^ B  -=-  cosec-  B  =  sin* B. 

8.  sec"  B  -  tan-*  B  =  sec-  B  +  tan^  B. 

9.  (sin  B  -  cos  f))-=  1-2  sin B  cos  B. 

10.  l-tan*^  =  2sec-^-sec*^. 

11.  l+S'^  =  (eosec B-\-cotB)\ 


(> 


12.  (sin^  +  o.ohB)--\-  (sin^  -  cos 5)-=  2 

13.  sin"^-cos"^=siir6'- cos-^. 

14.  sin- ^  +  vers- ^::=  2(1- cos  e). 

15.  cot'^e-cos2^  =  cot2dcos2d. 

EXAMPLES. 
In  a  right  triangle  Al^C  (see  tignre  of  Art.  15)  given 
1.    ((  =  P'  +  ptji  c  =  <f+2^Q ;  calculate  cot  A 


Ans.  :^lt=jt. 
P 


2.    b  =  Im  -{r  n,  c  =  /?i  -?-  m ;  calculate  cosec  A. 


n' 


Vx"  —  VI* 


EXAMPLES. 


45 


3 

3.  sin  A  =  '-,  c  =  liO.o  j  calculate  a. 

5 

4.  Given  cot^A  =  tan  A ;  find  A. 

5.  "  sin  A    =cos2A;  find  A. 

6.  "  cot  A    =^  tan  G  A  ;  find  A. 

7.  "  tan  A    =oot8A;findA. 

8.  "  sin  2  A  =  cos  3  A  ;  find  A. 


Ans.  12.3. 


9. 
10. 

11. 

12. 
13. 

14. 
15. 

16. 

17. 

18. 
19. 

20. 


« 


C( 


u 


(( 


(( 


(( 


(t 


({ 


« 


a 


2 

sin  A    =  - :  find  cos  A  and  tan  A. 
3' 


"       cos  A    =-;  find  sin  A  and  tan  A 
5 


cosec  A  =  - ;  find  cos  A  and  tan  A. 
o 


1 


V5     J> 
3     3 

5'  r 

V7  _3_ 

I  Z*^     1 

sin  A    = — ;  find  cos  A  and  tan  A.  \-,    

V3  >'3'    V3 

cos  A    =  b;  find  tan  A  and  cosec  A. 

vn^'^      1 


sin  A    =.6;  find  cos  A  and  cot  A. 

tan  A    =  - ;  find  sin  A. 
5 

8     » 
cot  A    =  — ;  find  sec  A  and  sin  A. 

15 

12 
sin  A    = — ;  find  cos  A. 

13' 

cos  A    =  .28;  find  sin  A. 

tan  A   =-;  find  sin  A. 
3' 

sin  A    =-;  find  cos  A.   -»  ' 


VI  -  b' 

4    4 
5'    3 

4 

V41 

8    15 

17'  17 

6 
13 

.96 

.    4 

fV2. 


^   I' 

! 

I 


- 


46 


PLANE  TRIGONOMETRY. 


21.   Given  tanA  =  ,^;  lind  sinA  and  secA.      Atis.   -,    -• 


22. 


23. 


"      taiid  =-;  tind  sin^  and  cosd. 
b 


a 


(( 


COS0  =-;  find  sin^  and  cot 6. 
a 


6. 

■V¥^l        1 


24.   If  sin  ^  =  a,  and  tan  ^  =  6,  prove  that 
(l-a^)(l  +  &^)  =  l. 

Express  the  following  functions  in  terms  of  the  functions 
of  acute  angles  less  than  45° : 


25.  sin  168°,  sin 210°. 

26.  tan  125°,  tan  310°. 

27.  sec  244°,  cosec281°. 

28.  sec  930°,  cosec  (-600°). 

29.  cot  460°,  sec  299°. 

30.  tan  1400°,  cot  (-1400°). 


Ans.  sin  12°,  -sin 30° 

-cot 35°,  -cot 40° 

—  cosec  26°,  — secll° 

-sec 30°,  sec 30° 

-tan  10°,  cosec 29° 

-tan 40°,  cot 40° 


Find  the  values  of  the  following  functions : 

V3       1 


31.   sin  120°,  sin  135°,  sin  240°.       Ans. 


32.   cos  135°,  tan  300°,  cosec  300° 


V3 


V2' 


V2 
-V3, 


V3 


33.   sec315°,  cot330°,  tan 780° 


34.   sin  480°,  sin  495°,  sin  870° 


36.   tan  1020°.  sec  1395°,  sin  1486° 


V2,    -V3,    V3. 
V3      1      1 


2  '    V2    2 

-V3,  V2,  -K- 

V2 


EXAMPLES. 


47 


36.   sin  (-240°),  cot  (-675°),  cosec  (- 690°). 


Ans.  ^,   1,  2. 

37.  cos(-300°),  cot(-316°),  cosec  (- 1740°). 

1  *> 

1       ■" 

2\      va 

38.  tair^660°,  cos"  1020°.  -3V3,  -- 
Find  the  value  of  the  sine,  cosine,  and  tangent  of  the 


following  angles : 

39.  -300°. 

40.  - 135°. 

41.  750°. 

42.  -840°. 

43.  1020°. 

44.  (2n  +  l)^-|. 

45.  (2?i-l),r  +  v 


A        V3     1 
Alls.    -     ■x/.S 

'     ,      1 


TT^l 


V2         V2 
1     V3       1 


2'  2  '  V3 

_V3  _1  /- 

"2  '        2'  "^"^ 

V3  1  /- 

~T''  2'  "^^ 

V3     _1     _   /. 

2  '  2'  ^"^ 

1  V3  1 


2  '    V3 


Trove,  drawing  a  separate  figure  in  each  case,  that 

46.  sin340°=sin(-160°).  , 

47.  sin  (-40°)  =  sin  220°. 

48.  cos320°=-cos(140°). 
A9.  cos  (-380°)= -cos  560°. 
50.  cos  195°=  -  cos  ( -  15°). 
61.  cos380°=-cos660°. 


rr 


48 


PLANE   TRIGONOMETRY. 


1 

• 

;' 

!  ill  jiiiKj 
1^  41 


m 


>      I 


52.  •  cos  (  -  225°)  =  -  cos  ( -  45"). 

53.  cos  1()0«°  =.  -  cos  1 185°. 

54.  Draw  an  angle  whose  sine  is  -• 

55.  "       "       «  «      cosecant  is  2. 

56.  "       «       "  *      tangent  is  2. 

57.  Can  an  angle  be  drawn  whose  tangent  is  427  ? 


58. 
59. 


((       u 


U         <( 


((         (( 


((        <l 


« 


i( 


a 


cosine  IS  -  ? 
4 


"       secant  is  7  ? 


60.  Find  four  angles  between  zero  and  -f-  8  right  angles 
which  satisfy  the  equations 

(1)  sinA  =  sin20°,   (2)  sinA  =  -— ,   (3)  sinA  =  --. 
y  \  )  V2  7 

Ans.   (1)  gO°,  160°,  380°,  520°; 

fC)\     5ir        7v        13  TT        15  TT 

(3)  ^, 

61.  State  the  sign  of  the  sine,  cosine,  and  tangent  of  each 
of  the  following  am  les  : 


4' 

4  ' 

4   ' 

1377 

rr      ' 

I 

'>9  — 

Wii  TT 

7  ' 

27  TT 

7 

(1)  275°;   (2)  -91°;   (3)  -193°;  (4) 
(5)  -1000°;  (6)  2mr-\- 


350°; 


Sir 


+  ;  («)  +, -,  - 


Ans.  (1)  -,+,-;  (2)  -,-,+;  (3)  + 
(4)  +,+,+;  (5)  +,  +, 

Prove  the  following  identities : 

62.  (mi'd-\-Gos'e)-=l. 

63.  (sin^^  -  (iOH'eyr=  1  -  4cos=^^  -j-  4cos*d. 

64.  (sin e  +  cos  0)-=  1  +  2 sin  0  cos 0. 

65.  (sec^-tan6')(sec0  +  tau^)  =  l. 


411 


EXAMPLES. 


49 


66.  (cosec  $  —  cot  B)  (cosec  ^  +  cot  ^)  =  1.  * 

67.  aio!^d-\-Gos'^d  =  {suid-\-cosd){i-sm6cos$). 

68.  sin''^  +  cos«^  =  sm^0  +  cos*^-sin-^cos''^. 

69.  8iii=*  $  tan''  d  +  cos=^  0  cot^  6  =  tan-  6  +  cot'^  6-1. 

70.  sin ^ tan''^+  cosec^ sgc-6=  2 tan ^ sec^—  cosec ^4-  sin 0. 

71.  cos''^  — sin''d:=(cos^-sine)(l  +  sin^cosd).  , 

72.  sin«^  +  cos"^  =  l-asin2^cos2d. 

73.  tan«  +  tan/?  =  tanrttan/3(cot«4- tJot/8). 

74.  cot «  +  tan  jS  =  cot « tan  /3  (tan  «  +  cot  /8) . 

75.  1  —  sin«  =  (l +sin«)(sec«— tan«). 

76.  1  4-cos«  =  (1  —  cos«)(cosec«  — cot«).  » 

77.  (1  4-sin«  +  cos«)^=2(l +  sin«)(l -f  cos«). 

78.  (1  —  sin«  —  co8a)^(l  +  sin«+cos«)*=  4sin''«cos*«. 

79.  2  vers  a  —  vers^«  =  sin^t.  . ,  • 

80.  versa  (1  +  cos  a)  =  sinVx. 


50 


PLANE  TRIGONOMETRY. 


mt 

1 

' 

|nj 

'  i 

'  1 

)• 


m 


CHAPTER  III. 
TRIGONOMETEIO  FUNOTIONS  OF  TWO  ANGLES. 

42.  Fundamental  Formulee.  —  We  now  proceed  to  express 
the  trigonometric  functions  of  the  sum  and  difference  of 
two  angles  in  terms  of  the  trigonometric  functions  of  the 
angles  themselves. 

The  fundamental  formulix,  first  to  be  established  are  the 
following : 

«- sin  (a;  +  y)  =  sin  a;  cos  y  +  cos  x  sin  y  .  ...  {\\ 

cos(a  +  2/)  =  cosa;cos2/  — sinxsiny  .  .     .     .  (2) 

^sin  (a  — 2/)  =  sinajcost/  — cosiKsiny  .  .     .     .  (3) 

cos  (a;  — y)  =  cosxcosy -f  sinajs'ny  .  .     .     .  (4) 

NoTK.  — Here  x  and  y  are  iinglcs;  bo  that  {x  +  y)  and  {x  —  y)  are  also  angles. 
Hence,  Bin  (x  +  y)  is  the  Bine  pf  an  angle,  and  is  not  the  same  as  Bin  x  +  Bin  y. 
Sin  (X  +  2/)  is  a  single  fraction. 
Bin X  +  Bint/  is  the  sura  of  two  fractions. 

43.  To  prove  that 

sin  (a;  4-  y)  =  sin  x  cos  y-\-Q,osx  sin  y, 
and  cos  {x-\-y)  =  cos  x  cos  y  ~  sin  x  sin  y. 

Let  the  angle  AOB  =  x,  and  the 
angle  BOC  =  y  j  then  the  angle 
AOC  =  a;  +  y. 

In  OC,  the  bounding  line  of  the 
angle  (x-i-y),  take  any  point  P, 
and  draw  PD,  PE,  perpendicular 
to  0  A  and  OB,  respectively ;  draw 
EH,  EK,  perpendicular  to  PD  and  OA. 


!: 


FUNDAMENTAL  FORMULA. 


61 


Then  angle  Eril  =  mf-  REV  =  HEO  =  AOE  =  x, 

.    ,     .     ,      DP     EK  +  l'H     EK^PH 
8m(xH-2/)=— =  -———  =  -- +  -^ 


J  4^ 


OP      OP 

_ek  oe*.  ph  pe 
"oe'op    pe'op   '"  ••''" 

. "  s=  sin  a;  cos  2/ +  cos  a;  sin  y. 
,     .     ,     OD     OK -HE     OK      HE 


^OK  OE_HE  PE  ..<•>'•. 

■■  ""OE  'op      PE'OP  '^^^ 

' ;    .  '         =  cos  X  cos  y  —  sin  x  sin  y. 

Note.— These  two  formr.lw  have  been  obtained  by  a  construction  In  which  x  +  y 
is  an  acute  angle;  but  the  proof  is  perfectly  general,  and  applies  to  iingieH  of  any 
magnitude  whatever,  by  paying  due  regard  to  the  algebraic  signs.    For  example, 

Let  AOB  =  a;,  as  before,  and  BOC=y;      v  g 
then  AOC,  measured  in  the  positive  direc- 
tion. Is  the  angle  x  +  y. 

In  OC,  take  any  point  P,  »nd  draw  PD, 
PE,  perpendicular  to  O  A  and  OB  produced ; 
draw  Ed  and  EK  perpendicular  to  PD  and 
OA. 

Then,  angle  EOK  =  IW-a;; 

KPH==KOK  =  180°-x; 
and  COE=y-180°. 

DP         EK  -  PH 


sin  (x  +  y)  =  — 


OP 


OP 


EK^PH 
OP     OP 


EK    OE  ^  PH    PE 
' OE  *  OP     PE  '  OP 


=  -  sin  (180'^  -  X)  cos  (y  - 180°)  +  cos  (180°  -  x)  sin  (y  -  180°) 
=  sin  X  cos  y  +  cos  x  sin  y.  (Art.  36) 

'^   "'  OP    OP    OP  OP 

^OK  OE  EH  PE 
OE '  OP  >E  OP 


*  The  introduced  line  OE  is  the  only  line  in  the  figure  which  Is  at  once  a  side  of 
two  right  triangles  (OEK  and  OEP)  into  which  EK  and  OP  enter.  A  similar 
remark  applies  to  PB. 


r^'  1 


6J2  PLANE  TRIGONOMETRY. 

=  COB  (180°     X)  COB  (y  - 180")  +  sin  (180='  -  x)  sin  {y  - 180'*) 
=  C0BXC0B^  — Binxsiny.  (Art.  35) 

The  student  should  notice  that  the  words  of  the  two  proofs  are  very  nearly  the 


Mme. 


44.  To  prove  that 

sin  (x  —  y)  =  sin  x  cos  y  —  cos  x  sin  ?/, 
and  cos  (x  —  y)  =  cos  x  cos  y  +  s\nx  sin  y. 

Let  the  angle  AOB  be  denoted 
by  x,  and  COB  by  y ;  then  the  angle 
AOC  =  x-y. 

In  OC  take  any  point  P*,  and 
draw   PD,   PE,  perpendicular    to 
OA   and  OB   respectively;   draw 
EH,  EK,  perpendicular  to  PD  and  O 
OA  respectively. 

Then  the  angle  EPH  =  90°  -  HEP  =  BEH  =  AOB  =  x. 

-.  .     DP     EK  -  HP     EK     HP 

sin  {  X  —  v )  =  —  = =  — 

^        ^^     OP  OP  OP      OP 

^EK  (2E_HP  PE 

oe'op    PEOP 

=  sin  .13  cos  2/ —  cos  05  sin  y. 

noc^^     ,A     ^^I^     OK  +  EH      OK^EH 

cos  (a;  —  y)  =  —  = ' = 

^       -^^     OP  OP  OP      OP 

^OK  OE     EH  PE 
oe'op     PE   OP 

=  cos  X  cos  y  -f  sin  x  sin  y. 

Note  1.  — The  sign  in  the  ezprcBHion  of  the  sine  is  the  same  as  it  is  in  the  angle 
expanded ;  in  the  cosine  it  is  the  opposite. 

*  P  is  taken  In  the  lin«  boundtns  the  angle  undsr  consideration;  i.«.,  AOO. 


4VJ 


SINE  AND  COSINE. 


53 


NoTX  2.  —  In  this  proof  the  angle  x  -y  \is  acute ;  but  the  proof,  like  the  one  given 
in  Art.  43,  upplies  to  angles  of  any  magnitude  whatever.    For  example, 

Let  AOB,  measured  in  tlie  positive  x 

direction,   =x,  and   BOC  =  y.     Tlien  \ 

AOC  =  x-y  Hi      i.E 

In  OC  talce  any  point  P,  and  draw    O 
PD,  PE,  perpendicular  to  OA  and  OB 
produced:  draw  EH,  EK,  perpendicu- 
lar to  DP  and  AO  produced. 

Then, 

angle  EPH  =  EOK  =  AOB  =  360^-  a?, 
and      POE  =  180°  -  y. 


OP 


OP 


co»(x-y)  =  - 


^EK    OE_HP_PE 
OK    OP     PE  "  OP 

=  Bin  (360°  -  X)  cos  (180°  -y)-  coi  (SflO"  -  x)  sin  (180°  -  ^0 

=  (—  sin  3!)  ( -  cos  y)  —  cos  xainy 

=  sin  X  cos  y  —  cos  x  nin  y, 

OD         OK  +  HK 


OP 


OP 


OK    OE_HK_PE 
OE  '  OP     PEOP 


=  -  cos  (360°  -  X)  cos  (180°  -  y)  -  sin  (360°  -  x)  lin  (180°  -  y) 
=  (—  cos  X)  (—  cos  y)  -  (—  sin  x)  sin  y 
=  cos  X  cos  y  +  sin  X  sin  y. 

NoTK  3. —  The  four  fundnnientai  formulae  Just  proved  are  very  Important,  and 
must  be  committed  to  memory.  It  will  be  convenient  to  refer  to  them  as  the  '  x,y' 
formulie.  From  any  one  of  them,  all  the  others  can  be  deduced  in  the  following 
manner : 

Thus,  from  cos  (x  —  y)  to  deduce  sin  (x  +  y) .     We  have 
COS  (aj  — y)  =  cosa;cosi/  + sinajsiny .     .     .     .     (1) 

Substitute  90"—  x  for  x  in  (1),  and  it  becomes 
cos  1 90°-  (x  +  y)  I  =  cos  (90°-  x)  cosy  +  sin  (90°-  a;)  siny. 
.'.  sin  (a;  4-y  )  =  sina;cos2/  + cosajsiny.  (Art.  29) 

The  student  should  make  the  substitutions  indicated 
below,  and  satisfy  himself  that  the  corresponding  i  -ults 
follow : 


64 


PLANE  TRIGONOMETRY. 


m 


i 


I 


li 


u 


t( 


it 


(t 


From 

sin  (ic  +  y)  to  deduce  cos  {x  +  2/)  substitute  (90°+ a;)  for  x. 

ti          tt        «  cos{x  —  y)  "  (90° -a;)  for  a;. 

cos(a;  +  2/)  "        "  sin(x  +  y)  "  (90°  +  x)  ^ or  ». 

«  sin  (a; -2/)  "  (90°- »)  for  a;. 

«  cos(a;-y)  "  —yioxy. 

etc.                              etc.  etc. 

EXAMPLES. 

1.   To  find  the  value  of  sin  15°. 

sinl5°=sin(45°-30°) 

=  sin  45°  cos  30°  -  cos  45°  sin  30°     . 

V2'   2       V2'2 

^V3-l 

2V2 

V3  +  1. 
2V2  ' 

V3-1 

2V2  * 

V3  +  1 

2V2 


2.  Sh^w  that  sin  75°= 

3.  Show  that  cos  75°= 

4.  Show  that  cos  15°= 
3 


5 


5.  If   sina;  =  \   and   cosy  = 
cos  {x  —  y). 

6.  If   sin  X  =  -,    and    cos  y  =  .^ 
cos  (x—y). 


find    sin  (a;  -f  .v)    and 
.        63         ,  50 
^"*-   65'  ""^  65' 

find    sin  (a;  -f  y)    and 
Ans.  1,  and  -~  • 


IVH^  ^ 


FORMULAE  FOR   TRANSFORMATION. 


65 


45.  Formulae  for  the  Transformation  of  Sums  into  Prod- 
ucts. —  From  the  four  fundamental  formuUe  of  Arts.  43 
and  44  we  have,  by  addition  and  subtraction,  the  following: 

sin  (a;  4- 2/)  4- sin  (.1;  — 2/)  =  2 sin ic  cosy  .  .  .  (1) 

sin  (05  4-?/)  — sin  (a;  — y)  =  2  cos  a;  sin  2/  .  .  '.  (2) 

cos  {x  4-  y)  4-  cos  (a;  —  y)  =  2  cos  x  cos  ?/  .  .  .  (o) 

cos  (x  —  y)  —  cos  {x  4-  2/)  =  2  sin  x  sin  y  .  .  .  (4) 

These  formulae  are  useful  in  proving  identities  by  trans- 
forming products  into  terms  of  first  degree.  They  enable 
us,  when  read  from  right  to  left,  to  repla(^e  the  product  of  a 
sine  or  a  cofdne  into  a  sine  or  a  cosine  by  half  the  sum  or 
half  the  difference  of  two  such  ratios. 

Let  «4-2/  =  A,  and  x  —  y  =  li. 

.-.  a;  =  i(A  +  B),  and  y  =  ^(A-B). 

Substituting  these  values  in  the  above  formulae,  and 
putting,  for  the  sake  of  uniformity  of  notation,  x,  y  instead 
of  A,  B,  we  get 

sina;4- sin2/  =  2sin^  (a;4-2/)cos^(a;  — 2/)  .  .  (5) 

sin  X  —  sin  y  =  2 cos  ^  (a;  4-  y)  sin  ^{x  —  y)  .  .  (6) 

cosa;4-cos2/  =  2cos  }  (a;4-y)  cos^(.r  —  2/)  .  •  (7) 

C0S2/  — cosa;  =  2sin  ^  (a; +  2/)  sin^(a;  — y)  .  .  (8) 

The  formulae  are  of  great  importance  in  mathematical 
investigations  (especially  in  computations  by  logarithms)  ; 
they  enable  us  to  express  the  sum  or  the  difference  of  two 
sines  or  two  cosines  in  the  form  of  a  product.  The  student 
is  recommended  to  become  familiar  with  them,  and  to  com 
mit  the  following  enunciations  to  memory  : 

Of  any  two  angles,  the 

Sum  of  the  sines  =  2  sin  ^ sum  -cos^difP. 
Diff.  "    "       "     =2cosisum.sinidiff. 


~ 


T 


liii 


I 


U 


» 


66  PLANE  TRIGONOMETRY. 

Sum  of  the  cosines  =  2  cos  ^  sum  •  cos  J  diff. 
Diff.  "    "        "       =  2  sin  I  sum.  sin  ^  diff. 

EXAMPLES. 

1.  sin 5 a; COS 3 a;  =  ^- (sin 8a;  +  sin 2 a;). 

For,        sin5a;cos3a;  =  ^  |sin  (5a;  +  3a;)  +  sin  (5a;  — 3a;)| 

=  ^  (sin  8  a;  +  sin  2  x) . 

2.  Prove  sin ^  sin 3^        =^  (eos2e- cos4tf). 

3.  "      2sin^cos<^        =  sin  (^  +  <^)  +  sin  (^  — <^). 
2sin2^cos3<^  =  sin(2^-f  3<^)-f  sin(2^-3<^). 
sin  60° + sin  30° = 2  sin  45°  cos  15°. 
sin  40°  -  sin  10°  =  2  cos  25°  sin  15°. 
sinl0^+sin6^=  2sin8dcos2tf. 
sin  8  «  —  sin  4  a  =  2  cos  6  a  sin  2  a. 
sin3a;  +  sina;    =  2  sin  2  a;  cos  a;. 
sin3a;  — sina;    =  2  cos  2  a;  sin  a;, 
sin  4  a;  4-  sin  2  a;  =  2  sin  3  x  cos  a;. 

46.  Useful  Formnlee.  —  The  following  formulse,  which 
are  of  frequent  use,  may  be  deduced  by  taking  the  quotient 
of  each  pair  of  the  formulae  (5)  to  (8)  of  Art.  45  as  follows : 

1     sina;-}-  siny  _ 2 sin^(a;  -}-y)  cos^(a;  —  y) 
sina;— sin y     2cos^(a;-|- y)  sin^(x' —  y) 

=  tan  ^(a;  -|-  ?/)  cot  ^(a;  —  y) 

=  ^-^te±ll.  (Art.  24) 

tan^(x  — y) 

The  following  may  be  proved  by  the  student  in  a  similar 

manner : 

„    sin  X  -f  sin  y      .      i  /     ,     v 
2.    — -T '^  —  tan  ^  ( a;  -f  .?/ ) , 

cos  X  +  cos  y 


5. 

6. 

7. 

8. 

9. 

10. 

11. 

THE  TANGENT  OF  TWO  ANGLES.  67 

o    sin  a;  +  sin  V         1.1/  \ 

cos  y  —  cos  X 

cosaj  +  cosy  ' 

^    sin  X  —  sin  w       ^1. 1  /     ,     \ 
cos  2/  — cos  a; 

^     cos  X  4-  cos  V  1.  1  /      ,      \      i.  1  /  \ 

cos  2/  —  cos  x 

47.  The  Tangent  of  the  Sum  and  Difference  of  Two  Angles. 

—  Expressions  for  the  value  of  tan  (a*  +  y),  tan  {x  —  y),  etc., 
may  be  established  geometrically.  It  is  simpler,  however, 
to  deduce  them  from  the  formulae  already  established,  as 
follows : 

Dividing  the  first  of  the  'x,  y^  formulae  by  the  second, 
we  have,  by  Art.  23, 

tan  (X  +  y)=  ^^^-^tl)  =  sin  a;  cos  y  4- cos  a;  sin. y 
cos  (a; +  2/)      cosajcosy  —  sinajsiny 

Dividing  both  terms  of  the  fraction  by  cos  a;  cosy, 

sin  a;  cosy     cos  a;  sin  y 

.     .     .      cos  a;  cosy     cos  a;  cosy 

tan  (a;  -f-  y)  = '- : :— i 

cos  a;  cosy      sinajsiny 

cos  a;  cosy     cos  a;  cos  a; 

^  taji^JJany  ,^j.^  23)    .     .     .     (1) 
1  —  tan  X  tan  y 

In  the  same  manner  may  be  derived 

ta„(x-!,)  =  i?Sfn^ (2) 

1  -f  tan  X  tan  y 

Also,   cot(x  +  j,)  =  55tf-«2^ (3) 

cotaj  +  coty 

and  cot  (a:  -  y)  =  ^-5t?-£5*y±l (4) 

coty  —  cotx 


4 


1  !■* 


Hi'        \\ 


58  PLANE  rUlGONOMETRY. 

EXERCISES. 

Prove  the  following : 

1.  tan  (a;  +  45°)  =^^"^"^^- 

1  — tanaj 

tan  x  —  1 


2. 
3. 

4. 


tan  (x  -  45°)  = 


1  4-  tan  X 


sin  (x  4-  y)  _  tan  x  -f-  tan  y 
sin  {x  —  2/)         tan  a;  —  tan  y 

cos  (a;  —  y)  __  tan  a;  tan  y-\-l 
cos  (» -f  y)         1  —  tan  X  tan  y 


6.   sin  (a;  +  y)  sin  (x  —  y)  =  sin'' a;  —  sin''3/ 

=  co&'y  —  cos' a;. 
6.   cos  (a;  -f  y)  cos  (a;  —  y)  =  cos*  a;  —  sin'''y 


=  cos'y  —  sin'a: 

7. 

tan«±tan2/=«i'li^2/). 
cos  a;  cosy 

8. 

cota.±coty  =«^"(2/±^). 
smajsiny 

9. 

sin  2  a;     cos  2a; 

— : =  sec  a;. 

sma; 


cos  a; 


10.  If  tan  X  =  ^  and  tany  =  \,  prove  that  tan  (a;  +  y)  =  f, 
and  tan  (x  —  y)  =  |. 

11.  Prove  that  tan  15°  =  2  -  V3. 

12.  If  tan  x=  |,  and  tan  y  =  ^,  prove  that  tan (x  -f  y)  =  1. 
What  is  (x  -f  y)  in  this  case  ? 

48.  FormulsB  for  the  Sum  of  Three  or  More  Angles.  —  Let 
X,  y,  z  be  any  three  angles ;  we  have  by  Art.  43, 


EXAMPLES. 


69 


sin  (x  +  y  +  z)  =  sin  (x  -\-  y)  cos  z  -f  cos  (a;  +  y)  sin  z 
=  sin  a;  cos  y  cos  z  +  cos  x  sin  y  cos  « 
+  cos  a;  cos  2/ sin  2  — sin  a;  sin  y  sin  2   .     .  (1) 

In  like  manner, 

cos  (a;  +  y +  «)  =  cosa;co8y  C0S2  —  sin  a;  sin  y  cos  z 

—  sin  a;  cos  2/ sin  2  — cos  a;  sin  y  sin  2   .     .  (2) 

Dividing  (1)  by  (2),  and  reducing  by  dividing  both       ms 
of  the  fraction  by  cos  x  cos  y  cos  2,  we  get 

tan(x+y+2)=  tana; +  tany  +  tan^-_^"na- tany  tan2      .3. 
1  —  tan  X  tan  y  —  tan  v  tan  2  —  tan  2  tan  x 


1 


EXAMPLES. 

1.  Prove  that  sin  a;  +  sin  y  +  sin  2  —  sin  (a:  +  y  +  2) 

=  4sin|(a;  +  y)  sm^(y  +  2)  sin ^(2  +  x). 

By  (6)  of  Art.  44  we  have 

sin  a;  — sin  {x-\-y  +  2)  =  — 2cos|(2a;-f  y -I-2)  sin^(2/ +  2), 
and  siny  4-sin2  =  2sin^(y  4- 2)cos^(?/ —  2). 

.-.  sin  a;  +  sin  2/ +  sin  2  —  sin  (a;  4- y  +  2) 

=2sin|(y+2)cos|(y— 2)-2cos|(2x-         :)s'm^{y+z) 
=2sin|(i/  +  2)|cos^(.V  — 2)-cosi  ''t  '  y  +  z)l 
=28in^(y  4- 2)  2sin^(aj  +  y)  sin |(a: -f  2) 
=4 sin i (a;  4- y)  sin ^(y  4- 2)  sin  1(2 -I- a;). 

Prove  the  following : 

2.  cos X  4-  cos y  4-  cos 2  +  cos  (a;  4-  y  4-  2) 

=  4 cos  ^ (y  4- 2)  «os  ^ (2 -I- a;)  cos  ^ (a;  4- y) • 


m 


i 


"I 


iiii 


;  ! 


'*  til 


60  PLANE  TRIGONOMETRY. 

3.  8iii(«-f  y  —  2)  =  sin  a;  cos  y  cos  2  +  cos  a;  sin  y  cos  2 

—  cos  X  cos  y  si»  2  +  sin  a;  sin  y  sin  2. 

4.  sina;  +  siny  —  sin2  —  sin  (aj-f- y  —  2) 

=  4  sin  ^  (a?  —  2)  sin  ^  (y  —  2)  sin  ^{x-}-  y) . 

6.   sin  (3/  —  2)  4-  sin  {z  —  x)-\-  sin  (x  —  y) 

+  48in^(.?/— 2)  sin ^(2  —  x)  sin  J  (a;  —  y)  =  0. 

49.  Functions  of  Double  /Ingles.  —  To  express  the  trigo- 
nometric functions  of  the  angle  2x  in  terms  of  those  of  the 
angle  x. 

Put  y  =  a;  in  (1)  of  Art.  42,  and  it  becomes 
sin  2  a;  =  sin  a;  cos  x  +  cos  x  sin  x, 
or  sin  2  a;  =  2  sin  a;  cos  a; (1) 

Put  y  =  a;  in  (2)  of  Art.  42,  and  it  becomes 

cos2a;  =  cos*a;  — sin'a; (2) 

=  1— 2sin-a; (3) 

or  =2cos''a;— 1. (4) 

Put  y  =  a;  in  (1)  and  (3)  of  Art.  47,  and  they  become 

taii2a;=    ^^^"^, (5) 

l-tan*a;  ^  ^ 

cot2a;  =  ^^*'^~^ (6) 

2cot«  ^  ^ 

Transposing  1  in  (4),  and  dividing  it  into  (1),  we  have 

.       "    ".r    ^^  tan  a?.     •    •    •     •    •    •    •    •    *     lil 

l  +  cos2a;  ^  ' 


EXAMPLES.  61 

Note.  —  These  ieven  formulsp  are  very  important.     The  •tudent  muit  notice  that 
X  it  nny  nnRJe,  and  therefore  these  formulee  will  be  true  whatever  we  put  for  x. 

Thus,  If  we  write  -  for  x,  we  got 

•ln«->2sin?eos£ (8) 

3        2  ^  ' 

cos  x==  COB* --sin*- (9) 

2         a 

or  «=l-2sln»-  =  2cos«?-l (10) 

and  so  on. 

EXAMPLES. 

Prove  the  following : 

1.  2cosec2x     =seca;coseca;.  .  .  .; . 

2.  _.5^??^  =sec2x. 
cosec  X  —  2 

o       2  tan  a;  .    „ 

3.    —     =sin2x. 

1  +  tan*  a; 

.     1  —  tan'a;  „ 

4. —     =  cos  2x.  — 

1  4-  tan'a; 

6.   tan  a; + cot  a;  =  2  cosec  2  a;. 

6.  cot  X  —  tan  x  =  2  cot  2  a;. 

7        sin  a;  .      x 

7. =tan-' 

1  4- cos  a;  2 

Q        sina;  .x 

8.  ;; =  cot-« 

1  —cos  a;  2 

9.  Given  sin  45**=-^;  find  tan  22^**.         Ans.  V2-1. 

V2 

10.   Given  tan  a;  =  - :  find  tan  2  a;,  and  sin  2  a;.  — ,    — 

4'  7'    25 

50.  To  Express  the  Functions  of  3  a;  in  Terms  of  the  Func- 
tions of  X. 

Put  y  =  2a;in  (1)  of  Art.  42,  and  it  becomes 

V 


■i  '. 


62 


PLANE  TRIGONOMETRY. 


'ml 


8in3a5  =  siii  (2a; +  x) 

=  sin2a;co8a/  -f-  Q,os2xs\nx 

=  28ma;co82a;-f-(l  — 2sin*a;)sina;     (Art.  49) 

=  2sina;(l  —  8in''a;)  + 8ina;  — 2sin'a; 

=  38inx  — 48in''aj. 
cos 3a;  =  cos  (2x-f-x) 

=  cos  2  a;  cos  a;  —  sin  2  a;  sin  a; 

=  (2  cos' a;— 1)  cos  a;  — 2  sin*  a;  cos  a;     (Art.  49) 

=  4  cos' a;  — 3  cos  a;. 

tan3a;  =  tan  2a;  +  a; 

_  tan 2a;  +  tana; 
1  —  tan2a;  tanx 

2tana; 


1  —  tan' a; 


-  -f-  tan  X 


1  _    ^t3,n'a; 
1  —  tan' a; 

3  tan  a;  —  tan' a; 
1  —3  tan' a; 


or 


\ 


i- 


::! 


hi'li 


EXAMPLES. 


Prove  the  following : 


^     sin3a;     „       o     ■  ^ 

1.   — =  2cos2aj-fl. 

sin  a; 

o    sin  3a;  — sin  a; 

iS.    — =  tan  X. 

cos  3  a; -f- cos  a; 

Q    sin3a;4-cos3a;      o  •    o        -t 
3.  2__ =  28in2a;  — 1. 

cos  »  — sin  a; 


t        •''■■ 


FUNCTIONS  OF  TUB  UALF  ANGLE.  68 

4.   1 4- ! =  cot2x. 

tan  3x— tana;     cota;  — cot3a; 

6.   L=L^?^  =  (i4.2cosa;)«. 
1  —  cos  a; 

'  51.  Functions  of  Half  an  Angle.  —  To  express  the  f unc- 

tions  of  -  in  terms  of  the  functions  of  x. 
2 

Since  cosa;=  1  —  2sin*-, 

2  • 

or  =2co82~-l     ...    [Art.  49,  (10)] 

.   « a;      1  —  ros.T  ,^v 

•••  8»"  2  = ^ (1) 

and  cos2-  =  -^!l^ (2) 

^                      .    a;      ,      /I  —cosa;  ,ov 

Or  sin-  =  ±^ ^ (3) 

,                           a;       ,      /I  +  cos  X  ... 

and  cos-  =  ±-i/ — ^— (4) 

T>    J-  •  •        i.     a;       ,     /I— cosa;      ,  1— cosa;  ,t!\ 

By  division,  tan  -  =  ±  \\- =  ± .     .     (5) 

^  '        2         \l  +  cosa;  sina;  ^  ' 

By  formulsB  (8),  (4),  and  (5)   the  functions  of  half  an 
ngle  may  be  found  when  the  cosine  of  the  whole  angle  is 
given. 

52.  If  the  Cosine  of  an  Angle  be  given,  the  Sine  and  the 
Cosine  of  its  Half  are  each  Two-Valued. 

By  Art.  51,  each  value  of  cosa;  (nothing  else  being  known 
about  the  angle  x)  gives  \wo  values  each  for  sin-  and  cos-, 

one  positive  and  one  negative.    But  if  the  value  of  x  be 


64 


PLANE  TRIGONOMETRY. 


t 


ViTJii 


given,  we  know  the  quadrant  in  which  -  lies,  and  hence 

we  know  which  sign  is  to  be  taken. 

Thus,  if  X  lies  between  0°  and  360°,  ^  lies  between  0°  and 

180°,  and  therefore  sin-  is  positive;  but  if  x  lies  between 

360°  and  720°,  -   lies    between  180°  and  360°,  and  hen(!e 

sin-  is  negative.     Also,  if  x  lie  between  0°  and  180°,  cos- 

is  positive;    but  if  x  lie  between  180°  and  360°,  cos-  is 

negative. 

The  case  may  be  investigated  geometncally  thus : 

Let  OM  =  the  given  cosine  (radius  being  unity,  Art.  16), 

=  cos  X.     Through  M  draw  PQ  per- 
pendicular to  O A ;  and  draw  OP,  OQ. 

Then   all  angles  whose  cosines  are 

equal  to  cos  a;  are  terminated  either 

by  OP  or  OQ,  and  the  halves  of  these 

angles  are  terminated  by  the  dotted 

lines  Op,  Oq,  Or,  or  Os.     The  sines 

of  angles  ending  at  Op  and  Oq  are 

the  same,  and  equal  numerically  to 

those  of  angles  ending  at  Or  and  Os ;  but  in  the  former  ease 

they  are  positive,  and  in  the  latter,  negative ;    hence  we 

obtain  two,  and  only  two,  values  of  sin  -  from  a  given  value 
of  cos  X. 

Also,  the  cosines  of  angles  ending  at  Op  and  Os  are  the 
same,  and  have  the  positive  sign.  They  are  equal  numeri- 
cally to  the  cosines  of  the  angles  ending  at  Oq  and  Or,  but 
the  latter  are  negative ;  hence  we  obtain  two,  and  only  two, 

values  of  cos-  from  a  given  value  of  cos  a;. 

Also,  the  tangent  of  half  the  angle  whose  cosine  is  given 
is  two-valued.  This  follows  immediately  from  (5)  of  Art. 
fil. 


Th 


FUNCTIONS  OF  THE  HALF  ANGLE. 


66 


53.  If  the  Sine  of  an  Angle  be  given,  the  Sine  and  the 
Cosine  of  its  Half  are  each  Four-Valued. 


We  have 
and 

By  addition, 


2sin-co8-  =  8iiia5   . 
2       2 

snr  -  4- cos^  -  =  1     .     . 


sin  -  +  cos  -  ]  =  1  4-  sin  x. 

,      2  '-^ ' 


(Art.  49) 
(Art.  23) 


By  subtraction,  [  sin^  —  cos*-  |  =  1  —  sin 


X. 


sin- 4- cos- =  ±  Vl  -f  sin  a; 
2  2 


and 


•     X  X 

sm  —  cos-: 
2  2 


±  Vl 


sin  a; 


.-.  2sin^=  ±  Vl  +  8inx±  Vl  —  sina; 
2 


and 


2  cos 


X 


±  Vl  +  sina;  T  Vl  —  sina; 


(1) 
(2) 
(3) 

(4) 


Thus,  if  we  are  given  the  value  of  sina;  (nothing  else 
being  known  about  the  angle  x),  it  follows  from  (3)  and 

(4)  that  sin-  and  cosp  have  each  four  values  equal,  two 

by  two,  in  absolute  value,  but  of  contrary  signs. 
T/te  case  may  be  inventigated  geometrically  thus : 
Let  ON  =  the  given  sine   (radius  being  unity)  =  sina;. 

Through  N  draw  PQ  parallel  to  OA ;  b 

and  draw  OP,  OQ.    Then  all  angles 

whose   sines  are  equal  to  sina;  are 

terminated  either  by  OP  or  OQ,  and 


the  halves  of  these  angles  are  termi- 
nated by  the  dotted  lines  Op,  Oq,  Or,       '* 
or  0.9.     The  sines  of  angles  ending 
at  Op,  Oq,  Or,  and  Os  are  all  different  « 


B' 


rr 


66 


PLANE  TRIGONOMETRY. 


m 


'  ij 


I  ^'? 


in  value;  and  so  are  their  cosines.     Hsnce  we  obtain  four 

values  for  sin-,  and  four  also  for  cos-,  in  terms  of  x. 

When  the  angle  x  is  given,  there  is  no  ambiguity  in  the 

calculations ;  for  -  is  then  known,  and  therefore  the  signs 

and  relative  magnitudes  of  sin-  and  cos-  are  Known.    Then 

equations  (1)  and  (2),  which  should  always  be  used,  im- 
mediately determine  the  signs  to  be  taken  in  equations  (3) 
and  (4). 

Thus,  when  -  lies  between  —  45**  and  4-  45**,  cos-  >  sin -, 

and  is  ponitive. 

Therefore  (1)  is  positive,  and  (2)  is  negative'  and  hence 
(3)  and  (4)  become 


2sin--=  Vl  +  sina;—  Vl  —  sin  a?, 


2cos-  =  Vl  4-  sin  x  -f  Vl  —  sin  x. 


X 


X 


When  -  lies  between  45**  and  135°,  sir.      >cos-,  and  is 
2  '        ^         ,2 

positive. 

Therefore  (1)  md  (2)  are  both  positive;  and  hence  (3) 
and  (4)  become 

2  sin -'  =  Vl  +  sino;  -f  Vl  —  sina;, 


2cos-=  VI  -f-sina;—  VI  —  sina. 


And  so  on. 


54.  If  the  Tangent  of  9:1  Angle  be  given,  the  TangeDt 
of  it'  Half  is  Tvo- Valued. 


We  have 


tantf  = 


2  tan    ^ 


1-tan'^ 
2 


(Alt.  49) 


Bl'l  I! 


m 


FUNCTIONS  OF  THE  HALF  ANGLE, 


67 


Put  tan -  =  x\  thus 
2 

(1  -  ar')  tan  d=  2a;, 


a^H -xss  1. 

tautf 


.,  tan?  =  :c=Ill^^l-±^!-^. 
2     .  tantf 

Thus,  given  tan^,  we  find  tico  unequal  vahies  for  tan-, 
one  positive  and  one  negative. 

This  result  may  be  proved  geometrically,  an  exercise  which 
we  leave  for  the  student. 

55.  If  the  Sine  of  an  Angle  be  given,  the  Sine  of  One- 
Third  of  the  Angle  is  Three- Valued. 


We  have 
Put 


sin3x*  =  .'isinx  — 4sin'''a;    .     .     (Art.  50) 
jc  =  -,  and  we  get 

o 

•  8intf  =  38ln 4sin''-, 

:i  3       •         

a  cubic  equation,  which  therefore  has  three  :  ^obs. 


s 


i 


'4 


EXAMPLES.  ' 

1.   Determine  the  limits  between  which  A  must  lie  to 
satisfy  tue  equation 

28inA  =  — Vl  4-sin2A  —  VI  —  sin2A. 

By  (1)  and  (2)  of  Art.  63,  2  sin  A  can  have  this  value 
only  when 

sin  A  4-  cos  A  =  —  Vl  +  sin  2  A, 


aRil  sinA  — cosA  =  — Vl  —  sin2A; 

i.e.,  when  sin  A  >  cos  A  and  relative.        ; 


I  . 


68 


PLANE  TRiaONOMETR Y. • 


m'i 


Therefore  A  lies  between  225"*  and  .'U5°,  or  between  the 
angles  formed  by  adding  or  subtracting  any  multiple  of 
four  right  angles  to  each  of  these ;  i.e.,  A  lies  between 

2nir  +  —  and  2n7r  +  ^, 
4  4 

where  n  is  zero  or  any  positive  or  negative  integer. 

2.  Determine  the  limits  between  which  A  must  lie  to 
satisfy  the  equation 

2cosA  =  VI  +  sin2A  —  vl  —  sin  2  A. 

By  (1)  and  (2)  of  Art.  53,  2cosA  can  have  this  value 
only  v;hen 


cos  A  +  sin  A  =  VI  +  sin  2  A, 


and 


cos  A  —  sin  A  =  —  VI  —  sin  2  A 


i.e.,  when  sin  A  >  cos  A  and  positive. 
Therefore  A  lies  between 

2n7r-f  7  and  2n7r  +  ^, 
4  4 

where  n  is  any  positive  or  negative  integer. 

3.  State  the  signs  of  (sin  d -}- cos  0)  and  (sin  ^  — cos  6) 
when  $  has  the  following  values:  (I)  22°;  (2)  191°; 
(3)  290°;  (4)  345°;  (5)  -22°;  (0)  -275°;  (7)  -470°; 
(8)  1000°. 

Ana.  (1)  +,  -;  (2)  -,  +;  (II)  -,  _;  (4)  +,  -; 

(5)  +, -;  («)  +, +;  (7)  -, -;  (8).-,-^. 

4.  Prove   thjit  the    formul.T?   which  give  the  values  of 

sin  -  and  of  cos-  in  terms  of  sinx  are  unaltered  when  x 
•>  «> 

has  the  values 

( 1 )  92°,  268°,  900°,  4 htt  -f  | tt,  or  (4  n  +  2)  tt  -  |  tt  ; 

(2)  88°,  -  88°,  770°,  -770°,  or  4nir  ±  5. 

o 


VALUES  OF 

5.  Find  the  limits  betw» 
2  sin  A  =  Vl~-^ 

56.  Find  the  Values  o: 
(4),  aud  (5)  of  Art.  61, 

siu22i°=^' 


cos22i 


001° 


♦  -- 


OMETRY. 

of  36°. 

,    .     .    [(3)  of  Art.  49] 


,  and  cosine,  respec- 
'7T,  and  3x-=l08°.^  /^ 


ANGLES  OF  A   TRIANGLE. 


71 


We  have  A  +  B  +  C  =  180". 
.-.  8in(A-f-B)  =sinC,  and  sin— ^-  =  cos^-  (Arts.  15 and 30) 


,,  .    A  +  li  „  A-B 

'>  QUI ' COS  

2  2 


Now       8inA  +  sinB=2sm— ■'^— -cos^^^^-jY"^  .     (Art.  45) 


=  2 cos 5  cos  ^^^^^    .    .  (Art.  15) 
2            ^ 

and  8inC  =  2sin5^eos^  .     .    .    .  (Art.  49) 

2        2 

=  2  COS  ^-±^008^    .    .  (Art.  15) 
2            2 

,.•.  sin  A  4-  sinB  +  sinC  =  2cos|cos^~-  +2co8|cos^^ 

„       Of      A  -  B   .    ^  A  -f-  B\ 

=  2cos-f  cos — hcos — ^  J 

=  2cos^/'2cos|cos|y  (Art.  45) 

.ABC  ,1^ 

=  4cos-cos-cos-   .     .  .    .     (1) 

Z        M       Z 

Again,    cos  A  -f  cos  B  =  2  cos  -^  cos  ^—  .  (Art.  45) 

2  ^ 


and 


o  .   C       A-B 

c=2sm^-cos— ^; 

cosC  =  l-2sin'^^    ....     (Art.  49) 


.'.  cosA+  cosB  +  cosC=l  +  2sin-Yco3— j^ sin^ 

=  l  +  2sinf(oosAr-B_eosA±B) 

=  l+4sm^smj;jsm5      .    .     (2) 
2         <^         M 


1 


72 


PLANE  TRIGONOMETRY 


Again,     tan(A  +  B)  =  -tanC (Art.  30) 


_  U  nA  -f-  tan  B 


.     .     .     (Art.  47) 


1  —  tan  A  tan  13 
•••  tan  A  +  tanB  =  —  tan  C  (1  —  tan  A  tan  B). 
.  •.  tan  A  -f  tan  B  -f  tan  C  =  tan  A  tan  B  tan  C    .     .     .     .     (,*3) 

Note.  — The  ntuilcnt  will  observe  that  (1),  (2),  and  (3)  follow  dlreclly  from 
Kxttiiiplea  1  and  2,  and  formula  (3),  renpectivel^,  of  Art. 48,  by  puttiug 

A  +  B  +  C  =  180°. 


IH 


ill 


:t- 


EXAMPLES. 

Prove  the  following  statements  if  A  +  B  -f  C  =  180°: 

1.  co8(A  +  B-C)  =  -cos2C. 

ABC 

2.  sin  A  +  sin  B  —  sin  C  =  4  sin  -  sin  —  cos  -  • 

2        2        2 

3.  sin2A +  sin2B +  sin2C  =  4sin  A  sinB  sinC. 

4.  sin  2  A  -I-  sin  2  B  —  sin  2  C  =  4  sin  C  oos  A  cos  B. 

6.   tanZA  —  tan4A  —  tan3A  =  tan7  A  tan4  A  tan3A. 

6.  sinA  — sinB  +  sinC  =  4sin  — cos     sin  — 

2   2   2 

„       ,  A  ,   .B  ,   .C    .A  .B  ,C 

7.  cot  — h  cot  — f-  cot  -  =  cot  cot  cot  -- 

222  222 

8.  tan  A  —  cot  B  =  sec  A  cosec  B  cos  C. 

60.  Inverse  Trigonometric  Functions.  —  The  equation 
8in^  =  .T  means  thiit  6  is  the  angle  whose  sine  is  ic;  this 
may  be  written  d  =  sin^ir,  where  sin^x*  is  an  abbreviation 
for  the  angle  (or  arc)  tvhose  sine  is  x. 

So  the  symbols  cos  'x,  tan~'a;,  and  sec" \?/,  are  read  "the 
angle  (or  arc)  whose  cosine  is  a:,"  "  the  angle  (or  arc)  whose 
tangent  is  a;,"  and  "  the  angle  (or  arc)  whose  secant  is  y." 
These  angles  are  spoken  of  as  being  the  inverse  sine  ofx,  the 


INVERSE  TRIGONOMETUIC  FUNCTIONS. 


73 


inverse  cosine  of  x,  the  inverse  tangent  of  x,  and  the  inverse 
secant  ofy,  resjH'otively.  Such  expressions  are  called  inverse 
lri(jonumetric  functions. 

Note.  —The  Htudent  muat  bu  careful  to  notice  that  —  1  in  not  au  exponent,  8in''x 

Is  not  {•In  X)"',  which  = -i—. 
■Inx 

1  ^3  1 1 

Notice  bUo  that  Bin'*  —  =  coa'' -  la  not  uu  identity,  but  ia  true  ouiy  for  Ui«  jiur- 

tioulur  an({lu  )H)<^. 

This  Udtntion  ta  oniy  aniiloyou»  to  the  uiie  of  rxpuncnts  In  mnltlplioatlon,  where 
we  have  tt''<i  a"-  1.  TIiuh.  con  '  (coix)  jt,  and  iiln(atn''x)  x;  that  !•,  com'' 
Ih  inverse  to  com,  and  applied  t>t  It  annuls  It;  and  so  for  other  functions. 

The  French  method  of  writing  inverse  functions  is 
arc  sinx,  arc  cosx,  arc  tanx,  and  so  on. 


EXAMPLES. 

1.    Show  that  30"  is  one  value  of  sin'^. 
We  know  that  sin 30°=  ^.     .-.  30°  is  an  angle  whose  sine 
is  h',  or  30°=  sin  '^. 

L*.   Prove  that  tan"'  i  -|-  tan"'  \  =  45°. 
tan~'.}  is  one  of  the  angles  whose  tangent  is  ^,  and  tan~'^ 
is  one  of  the  angles  whose  tangent  is  ^. 


Let 


ct  =  tan"'  ^,     and     /3  =  tan~'  ^ ; 


then  tana  =  ^    and    tan  ft  =  ^. 

xNow      '  tan  («  +  )«)  =  i^»  « +-t^»A^ 

1  —  tan«taniC 


•         f         * 


:tan/3 

=  _i±i-.  =  i 

But       tan  45°  =  1,        .  •.  «  +  /8  =  45° ; 
that  is,         .  taH-'  ^  +  tan"'  |  =  45°. 

Therefore  45°  is  one  value  of  tan~U  4-  tau-*|. 


(Art.  47) 


74 


PL  A  Nt:  TltlGONOMETH  Y. 


3.   Prove  that  tau-'aj  +  tan-'w=:  tan-'~i^. 


Let 


Now 


tau~'  X  =  A.      .  •.  tan  A  =  ar. 
tan~'y  =  B.       .♦.  tanB  =  y. 


tan  (A -I- B) 


tan  A  4-  tanB 
1  —  tan  A  tan  B 

l-xy 


.-.  A  +  B  =  tan-'  ,^ti^. 
i-xy 

.-.  tan-'a?  +  tan-'«  =  tan"' ^'^^  . 

1-xy 

Any  relations  which  have  been  established  among  the 
trigonometric  functions  may  be  expressed  by  means  of  the 
inverse  notation.     Thus,  we  know  that 


4.  cos  X  =  Vl  —  sink's;. 

This  may  be  written         a;  =  cos"' Vl  —  siu^a;    .     .     (1) 

Put  8inar  =  ^;   then  «  =  sin-'^. 


Thus  (1)  becomes      sin"'  $  =  cos*'  Vl  —  ^. 

6.   By  Art.  49,  cos2d  =  2cos'tf  -  1, 

which  may  be  written        26  =  coS-'(2co8-tf  —  1). 
Put  cos^  =  a?.     .-.  2cos-'iC  =  cos-*(2x'  — 1). 

6.   By  Art.  49,  sin  2 ^  =  2 sin tf  cos  9, 

which  may  be  written        2d  =  sin"' (2 sin tf  cos d). 


o 
u 


Put    sin^  =  x.     .-.  2siu"'a;  =  sin-'(2a;Vl  —  «')• 


TABLE  OF  USEFUL  FORMULAE. 


76 


7.   Prove 
8. 


sin-'a-scos"' Vl  —  a:*  =  tan-'       *     z' 


(( 


tan-'a:  =  sin-' 


X 


=  COS"' 


9.  « 

10.  " 

11.  « 

12.  « 

13.  " 

14.  '* 


Vi-f«* 

2tan-'x  =  tan-'-^^.     v^^ 
1  —  «* 


Vi  +  j;' 


sin  (2 sin-' x)  =  2x  Vl  —  x*.  " 

tan-'-  +  tan-'-  =  -. 
7  G     4 

•     co8-'-  +  28in-'?=120*.  ; 
2  2 

cot-'  3  +  cosec-'  VC  =  -• 

4 

Ssin-'a;  =  sin-'(3a;  ~  4.r') 


61.  Table  of  Usefttl  FormulsB.  —  The  following  is  a  list 
uf  important  formulae  proved  in  this  chapter,  and  summed 
up  lor  the  convenience  of  the  student : 

1.  sin  (a?  -\-y)    =  sin x  cosy  +  cos x  sin y  .     .     (Art.  43) 

2.  co8(x  +  y)    =cosxcosy  —  sinxsiny. 

3.  sin(x  — y)    =  sin x cosy —  cos xsiny   .     .     (Art.  44) 

4.  cos(x  — y)    =  cos  T  cos  y-f  sin  xsiny. 

5.  2 sin X  cosy  =  sin  (x -f  y) -f  sin  (x  —  y)     .     (Ar«.  45) 

6.  2cosxsiny  =  sin  (x  +  y)  — sin  (x  —  y). 

7.  2cosxcosy  =cos  (x  +  y)4- cos  (x  — y). 

8.  2  sin  X  sin  •  =cos(x  — y)  — cos(x-fy). 

9.  sinx-H  siny  =  28in^(.r  +  y)cos4^(x  —  y). 
10.  sinx  — iiny  =  2co.s^(x +  y)  si;,^(x  —  y). 


m 


\ 


■i 


i7   ' 


76 


PLANE  TRiaONOMETRY. 


11.  cosx  +  cosy  =  2co8  J(x-|- 1/)  co8^(aj  — y). 

12.  cosy —  cos a;  =  28iiii(ic  +  y)  sini(a;  —  y). 

13.  8nix-t8iny^tanHar  +  .v). (^,t.  4c) 

8111 JB  — amy      tan^vx  — y) 

14.  tan(x  +  y)   =  tana^  +  tan;^ ^^rt.  47) 

1  —  taiia;tany 

^  r    i.      /  V         tan  X  —  tan  y 

15.  tau  (a;  —  y)   =    ■  '^  • 

1  -f  tan  X  tau  y 

-«        i.  /     .    '        cotxcoty  — 1 

16.  cot  (x-fy>    =  —     ^ •• 

cotaj-f  coty 

17.  cot(x-y)    :^g2t^coty-Hl. 

cot»  — coty 

18.  tan(a;±4o'')  =  *-^-5-^^ (Art.  47) 

^  -^      tan  a;  ±  1  ^  ' 

19.  sin  {x-\-  y)  sin  (a;  —  y)  =  sin'' a;  —  sin'y  =  cos^y  —  cos'*. 

20.  cos  (x4-  y)  cos  (x  —  y)  =  cos*x  —  sin'y  =  cos^y  —  sin*x. 

21.  tanx±tany  =  ?ilL(^±J^. 

coax  cosy 


22.   cotx±coty=:»i-"Cy^^>. 

sin  X  sin  y 


23.   sin2x  =  2sinxcosx  = 


2tanx 


.    .     .     (Art.  49) 


1  -f  tan'x 

24.  co82x  =  cos'x  —  sin^'x  =  1  —  2sin''x  =  2cos'x  —  1 

1  —  tan'x 
1  -f  taii'x 

Of    1  —  oor2x      28in'x     .     « 
26.    ^  ^    — =!  tan*x. 

1  +eo8  2x      2co8'x 


SXAMPLES. 


77 


26.  tan  2  a;  = 

27.  cot2a5  = 


2tan« 

I    I  II        !■!■■■      » 

1  —  ttin'x 
voM^x  —  1 


2cotic 

28.  sin.Sjj  ssiiHina;  — •Isin'j! 

29.  cos  ti  a;  =  i  cob"'  x  —  3  cos  a;. 
3  tana;  —  tan"  a? 


30.   tan  3  a;  = 


l-3tan«x 


32.  co8»?  =l±i^. 

2  2 

33.  tan-'  x  +  tan"'  y  =  tan"'  :^^ 


.    .     (Art.  nO) 


31.   8in»-   =t^"*^"^^ \     .     (Art.  r.l) 

2  2  ^ 


•         •         • 


(Art.  6(») 


and  sin  (a  — /3). 


EXAMPLES. 

12 

1.  If  sin  a  = ; ,  and  sin  /3  =  -,  find  a  value  for  sin  ( « -f  ;8)» 

Ans.   ^^1^2      VO-4V2. 
9  '  9 

2.  If  cos  rt  =  r>  ainl  cos  B  —  —-,  find  a  value  for  -sin  ( «  -f  fl ) , 

o  41 

and  cos  («  +  /?).  :  ,  .        IfHi     l.'!;5 

205     205 

3  2 

3.  If  cos  «  =  *-,  and  c^s  B  =  -,  find  a  value  for  sin  (a  4-  j8), 

4  5 


and  8in(«  — /3). 


vln."».  - 


2V7  4-3V21     2V7-3V21 


20 


20 


4.    If  sin  «  =  -,  and  sin  B  =  -,  find  a  value  for  sin  {u  4-  fi), 
5  5 


and  co8.(rt4-/3). 


.^ns.  1, 


!^4 
25 


m 


IMAGE  EVALUATION 
TEST  TARGET  (MT-3) 


1.0 


I.I 


1.25 


12.5 


M    112.2 


.^~  m 


2.0 


^ 

1.4    III  1.6 


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^i$ 


78 


PLANE   TRIGONOMETRY. 


|i  iHiii 


i  Ml 


li:.'*!- 


)4 


I 


> 


'flr 


\ 


!§![ 


5.  If  sin«  =.6,  and  sinyS  =  — ,  find  a  value  for  sin(«— /8), 
and  cos  (a +  ^)-  ,16     33 

-^^"-   65'    65' 

6.  If  sin«=— -,  and  sinfl  = ,  show  that  one  value 

V5      ,       .,  VlO  .-.■-^•■■•;: 

of  «  +  j8is  45°.  ' 

7.  Prove  cos ^  +  cos 3 ^   =2eos2dcos0. 

8.  "      2cosacos/3       =cos  («-/?)  + cos  (« 4-/3). 
2sin3ecos5^  =sm8d-sin2^. 


9. 

10. 

11. 
12. 


a 


"      2cos|^cos-     =cos^  +  cos2^. 

"      sin4dsin^        =  |(cos3d  —  cos5^). 

"      2cosl0°sin50°=sin60^-f  sin  40". 


13.   Simplify  2cos2^cos^-2sin4esin^. 

Ans.  2  cos  3^  cos  2^. 


14.  Simplify  sm  —  cos  -  —  sin  -r-  cos  —  • 

z      z  z       z 

Prove  the  following  statements  : 

,t  : 

! 

15.  cosSrt  — cos7a  =  2sin5«sin2«. 

16.  sin  60°+  sin  20°=  2  sin  40°  cos  20°. 

17.  sin3^4-sin5^  =2sin4^cos0. 

18.  sin  7  0  —  sin  5 ^  =  2  cos  6  ^  sin  d. 

19.  cos50  +  cos9^  =  2cos7^cos2d. 

20.  «^"^^  +  ^"^^    =tan^. 
costf  +  cos2e  2 

21.  cos  (60°  +  A)  +  cos  (60°  -  A)  =  cos  A. 


cos  4  tf  sin  2^. 


w 


EXAMPLES. 


79 


22.  cos(45°-f- A)  +  cos(4.r- A)=  V2cosA. 

23.  sin(45°+ A)-siu(45°- A)  =  V2sinA. 

24.  cos2^  +  cos4^  =  2cos3^cos0. 

25.  cos4  6  — cos6^  =  2sin5^sin(?. 

26.  cos^  +  cos3^  +  cos5^  +  cos76  =  4cos0cos2dcos4^. 

27.  cot«  +  tany8  =^(«-^). 

sin«  cos^S 


.         1.      /D       cos  (« 4-  )8) 
cot  a  —  tan  /8   =  -^— ^^ — '-J^  < 

sin  a  cos /8 


28. 

29.    sin  (A -45°) 


sin  A  —  cos  A 


30. 
31. 

32. 

33. 

34. 
35. 
36. 

37. 

38. 

39. 


V2  sin  (A  +  45°)  =  sin  A  -f  cos  A. 

cos  (A  +  45°)  +  sin  (A  -  45°)  =  0. 

tan  (g  — <^)4-  tan<^   _  ^^^  ^ 
1  —  tan  {6  —  ^)  tan  <^ 

tan  (^  +  <^)  4-  tan  (^    _  ^^^  , 
l  +  tan(^+<^)  tan^ 


cos 


(^  +  <^)  -  sin  (^  -  <^)  =  2  sin /"- -  ^"j  cos 


0 


sin  7i6  cos  d  +  cos  n6  sin  ^  =  sin  (n  +  1 )  d. 

\^       4/      1-cot^ 

4- cot /'^  +  ?^  =  0. 

cot  r^  -  ^)  +  tan  (^  +  ^)  =  0. 

tan(n  +  l)<^-tan??0  ^  ^^^^  , 
1  +  tan  (n-{-l)  (f)  tan n</> 


*'"(*~i) 


■^  ■■,,*■ 


80 


PLANE   TRIGONOMETRY. 


If 


mmi 


40.    If  tanx  =  1,  and  tany  =  — -,  prove  that 

V3 


tan  {x -\- ij)  =  2 -\-  V3. 


41.    If  tana 


m  1 

,  and  tan/S  =  - — -7-^,  prove  that 


m  -f- 1  2  m  +  1 

tan(a  +  /3)=l. 

42.    If  tan  «  =  m,  and  tan  /?  =  ?t,  prove  that 

1  ~  mn 


cos  («  + 18)  = 


V(l+m2)(l+rO     ' 
43.    If  tan^  =  (a  -f- 1),  and  tan  ^  =  (a  —  1),  prove  that 


T».' 


Prove  the  following  statements  : 

44.  cos  {x  —  y-\-z)  =  cos  a;  cos y  cos 2  + cos ic  siny  sing 

—  sin  03  cos  y  sin  2;  +  sin  a;  sin  ?/ cos  «. 

45.  sin  («  —  1/ —  2)  =  sin.r  +  siny  4- sing 

-f  4sin^(.r— ?/)  sini(.r— «)  sin^(//-f  2). 

46.  sin  (a;  + y  —  2)  +  sin  (ic -|- 2:  — y)  +  sin  (?/  + 2  —  .t) 

=  sin  (a;  +  y  +  2!)  +  4  sin  a;  sin  2/ sin  2. 

47.  sin2a;4- sin2?/ +  sin22— sin2  (a; -f  y +2) 

=  4 sin  {x  +  ?/)  sin  (?y  +  2)  sin  (2  -f-  a?)- 

48.  cos 2 a;  +  cos 2?/ 4- cos 22  +  cos 2  (a;4- y +  2)       ' 

=  4  cos  {x  -f  y)  cos  (2/  +  2)  cos  («  +  aj) . 

49.  cos  {x  +  y  —  2)  4-  cos  (y  +  2  —  x)  +  cos  (2  +  a;  —  y) 

+  cos  (a;  +  y  +  2)  =  4  cos  a;  cos  y  C0S2. 


EXAMPLES. 


81 


50.    sin^a;  sin' 2/  +  siirz  +  sin^ (x^y  +  z) 

=  2|1  —  cos  (x  +  y)  cos  (y+z)  cos  (a+x)  |. 

51.    cos^  +  cos^y  +  cos^a;  +  cos^  {x-\-y  —  z) 

=  2|  1  +  cos  (a;  +  y)  cos  (a;— 2)  cos  (2/— 2;)  |. 
52.    cr'6x  sm(y  —  %)  -\-  cos y  sm{z—x)-{-  cos zsin(x—y)  =  0. 
sina;sin(?/  —  z)  +  sin  y  sin  (2— re)  +  sin2sin(x  — y)  =  0. 

^64.    cos  (a;  +  2/)  cos  (x  —  ?/)  +  sin  (y  +  2)  sin  (?/  —  «) 
—  cc  s  <  ^  -I-  2)  cos  (a;  —  2)  =  0. 


55. 


2-se.c^g 
sec^^ 


=  cos  2 1;. 


56.    cos2d(l-tan2e)  =  cos2^. 


57.    cot2^  = 


68.    sec2^  = 


2cotd 

cot^^  +  l 
cot^^-l* 


59. 


(-1 


+ 


6V 
cos  -  1  =  1  +  sin  0. 

2 


6 


60.    f  sin-  — cos- 1  =  1  —  sin^. 
2 


61. 


62. 


63. 


2 


14-secg 
sec^ 

cos  20 


00 


=  2cos^  - 
2 

1  -  tan  0 


l  +  sin20      l  +  tan0 


COS0 

l-sin0 


1  +  tan  - 
l-tan| 


y^ 


:-;  'r 


■  i   < 


82 


PLANE  TRIGONOMETRY, 


!■•'■? 
■.-.It 


%    I 


:t  ;. 


1,1 


ih  f 


64. 
65. 

m. 

67. 

68. 

69, 

70. 

71. 

72. 
73. 

74. 


1  +  sm  a;  +  cos  x         ,x 

T— ' — ^ ■ =  cot-- 

1  -f-  sin  X  —  cos  X  2 


cos^a;4-  sin^a; 
cosa;  +  sinic 

cos'^a;  — sin^x 
cos  a?  —  sin  a; 

cos*^  —  sin*^ 

sin3g     cos3g 
sin  6        cos  6 

cos3^  ,  sin30 


2  —  sin  2  a; 

2 

2  +  sin  2  35 


=  cos  2  6. 
=  2. 


sin  6 

sin4^_ 
sin2d" 

12 


+ 


sm 


cosd 
2  cos  2d. 

57r 


=  2cot2d. 


cos 


12 


sm- 


TT 

12 


TT 

cos  — 
12 


=  2V3. 


tan  (45°  +  a;)  -  tan  (45°  -  a;)  =  2  tan  2  x. 

tan  (45°  -  «)  +  cot  ( 45°  -x)  =  2  sec  2  a;. 

tan2(45°4-«)-l       •    o 

— -^^ — ^ =sm2x.      ,       ;    , 

tan2(45°+ic)  +  l  ,  ,  .       ,    , 

cos  (x  -f  45°)  o        i-     o 
\—^ — —-i-  =  sec  2  a;  — tan  2  a;. 

cos  {x  —  45  ) 


76. 

76.  tan  x  = 

77.  tan  x  — 
78. 


sinx-f  sin2a5 

I—     ■■■■    '■   '  ■      .1—  ■  • 

1  4-  cos  a;-}-  cos  2  a; 

sin  2  a;  — sin  a; 
1  —  cosaj  +  cos  2  a; 


cos  3  a; 


=  2  cos  2  a;  —  1. 


cos  a; 


X. 


i  iMi^:if 


r 


EXAMPLES. 


83 


\ 


r,n    3  sin  a;  —  sin  3  a:      .     , 

79. .^ =  tau^  X, 

cos3x  +  3  cos  a; 

cot'^'ic  —  3  cot  a; 


80.    cot  3  a;  = 


3cot2.'«-l 


81.    J-~^Q^^^=(l  +  2cosa;)l 
1  — cosa; 


Qo     sin  x  + cos  a;     1.0.0 
82.    — — I— ^ —  =  tan  2  a; -f  sec  2  a;, 
cos  X  —  sin  X 

go     cos  2  a;  +  cos  12  a;     cos  7  a;  —  cos  3  a;     2sin4a;     ^ 
cos 6 a;  4- cos 8 a;        cosa;  — cos3a;        sin2x 

84.  sin2a; sin22/  =  sin^(a;  +  y)  —  sin^(a;  —  y). 

85.  tan50°+cot50°=2secl0°. 

86.  sin  3  a;  =  4  sin  a;  sin  (60"  +  a;)  sin  (60°  —x). 

87.  cot|-tan-  =  2. 

8  8 

88.  tan4g=:^^^"^(^-^^^'^). 

l-6tan2^  +  tan*d 


89.  2cos^=V2  +  V2. 

90.  (3sin^  -  4siiv''^)2+  {4.cos'6  -  ScosOy  =  1. 


91. 


sin  2  g  cos  ^  9 

(1  +  cos  2  ^)  (1  +  cos  0)  ~  ^^2 


12  1 

92.  If  tan  0  =  -,  and  tan  <^  =  — ,  prove  tan  (2  ^  +  <^)  =  -• 

93.  Prove  that  tan-   and   cot-    are  the   roots   of  the 
equation 


Q?  —  2x  cosec  ^4-1  =  0. 


:  ■■; ' '; 


t 


r 


it 


84 


PLANE  TRIGONOMETRY. 


i'1 


94.    If  tan0=  -,  prove  that 
a 


cosO 


+  b      Vcos2e 

95.    Find  the  values  of  (1)  sin  9°,  (2)  008  9°,  (3)  sin  81°, 
(4)  cos  189°,  (5)  tan202i°,  (6)  tan97i°. 


Ans.  (1)  UV3+V§-^5-V5), 

(2)  I  (V3+ V5  +  V5  -  V5), 

(3)  sin  81°=  cos  9°, 

(4)  cosl89°=-cos9°, 

(5)  V2-I,  'r  ^    :^--^--  "^^■.;; 

(6)  _(V3+V2)(V2  +  1). 
96.   If  A  =  200°,  prove  that  •,.:.« 

(1)  2sin^  =+ Vl4-sinA  +  VI -sin  A. 


(2)    ta„|  =  ^±^'-±^^*-^- 
^   ^  2  tan  A 

97.   If  A  lies  between  270°  and  360°,  prove  that 
A 


(1)  2sin— =  +  Vl  -sin  A—  Vl  +  sinA. 

(2)  tan  -  =  —  cot  A  +  cosec  A. 

98.   If  A  lies  between  450°  and  630°,  prove  that 
(1)  2sin—  =  - Vl  +  sinA  —  Vl  -sin A. 


■>;.:^ 


J- 


(2)  2 cos 


Vl  +  sin  A  4-  Vl  —  sin  A. 


Ji  li  \ 


EXAMPLES. 


85 


Prove  the  following  statements,  A,  B,  C  being  the  angles 
of  a  triangle. 

99.    ^'°f -"'»-!  =  tang  tan ^—g. 

Sin  A  +sinB  2  *2 


sin3B-sin3C  _  tan 3  A 
cos3C-cos3B~      2 

A 

2      "2   ■  ^'"2 


100. 


101.    sin  ^  cos  ^  +  sin  I  cos  ^+  sin  ^  cos? 

9.  '}.  9.  9  9  O 


''''""'''''''  A       B       C    ' 
=  2cos  —  cos  —  cos  -  • 
2        2        2 

102.  cos'' I  +  cos''?-  -  cos^^  =  2  cos  A  cos?  sin?. 

2        ,2,.  2  2        2       2 

103.  sin  A  cos  A  —  sin  B  cos  B  +  sin  C  cos  C 

=  2cos  A  sinB  cosC.  ;^ 

104.  cos2A  +  cos2B  +  cos2C  =  -l -4cosAcosBcosC. 

105.  sin2A-sin2B  +  sin2C  =  2sinAcosBsinC.       V '^ 

106.  tan  I  tan  ?  +  tan  ?  tan  ^  +  tan  -  tan  ?  =  1.        '  - ' 

2         2  2        2  2        2 

Prove  the  following  statements  when  we  take  for  sin"^, 
COS"',  etc.,  their  least  positive  value. 

107.  sin-i^  =  cos-^^  =  cot-V3. 

108.  2tan-Hcos2^)  =  tan-'/'^5J^!^7-^^V       :        ' 


109.    4tan- 


-ll_tan-iA__![. 


5 


239     4 


110.    sin-i^H-sin-'l4-sin-'^  =  ^. 
5  17  85     2 


rr 


8(5  PLANE  TlilGONOMETRy. 

111.  tan-^  V5  (2  -  V3)  -  cot"'  V5  (2  +  Va)  =  cot-'  V5. 

112.  sec-V3  =  2cot-W2.  ,v 


113.    2cot-^a;  =  (!Osec 


2x 


\/3-fV2      .      -1     3      Sit 
V3-V2  ^^       * 

115.  sin-^-i=  +  cot->3  =  '- 

116.  cos-^|i^  +  2tan-'p  =  sm-^-.  :.; 

117     Ue  =  sin-'-,  and  <^  =  cos-^|  then  ^  +  </>  =  90°. 
5  o 

1  —  7? 
118.    Prove  that  cos  (2 tan"' a;)  =  ^"+^2* 


<,'.^  ■,  ■;* 


119. 
120. 

121. 
122. 


«        «    tan-^-  +  cosec-^\/lO  =  |-    ^  >  ,;^ 


<( 


(( 


(( 


2  i5       •  _i33 

"    2tan-'|-cosec-'-  =  sm    — •         -a^i 

"     2tan-i^  +  cos-*-  =  ^- 

"    sin-'  (cos  a;)  +  cos"'  (sin  y) -\- x -[- y  =  tt. 


123.        «        "     tan-'r-^+tan-'^-^  +  tan-'5^=/'7r. 


1  +  ^ 


124. 
126.       « 


«        «     tan-'— "-4-tan 


1-a- 


TT 


a; 


«     sin-^a  — sin-^2/ 


=  ?i7r  + 
2a;-l  4 


=  cos-'  {xy  ±  Vl-a;'-2/'  +  ^"V)- 


K:.„.  i  I 


NA.TUHE  AND   USE  OF  LOaASITIIMS. 


87 


■y 


/ 


CHAPTER   IV. 


/  LOGARITHMS  AND  LOGAEITHMIO  TABLES.  —  TEIQO- 
/  NOMETEIO  TABLES. 

62.  Nature  and  Use  of  Logarithms.  —  The  numerical  cal- 
culations which  occur  in  Trigonometry  are  very  much 
abbreviated  by  the  aid  of  logarithms ;  and  thus  it  is  neces- 
sary to  explain  the  nature  and  use  of  logarithms,  and  the 
manner  of  calculating  them. 

The  logarithm  of  a  number  to  a  giver  base  is  the  exponent 
of  the  power  to  which  the  base  must  \)e  ra'sed  to  give  the 
number. 

Thus,  if  a' =  711,  x  is  called  the  "  logarithm  of  m  to  the 
base  a,"  and  is  usually  written  «  =  log„?>i,  the  base  being 
put  as  a  suffix.* 

The  relation  between  the  base,  logarithm,  and  number  is 
expressed  by  the  equation,  '    •  ,v     ' 

■.      '        (base)'''^  =  number. 

Thus,  if  the  base  of  a  system  of  logarithms  is  2,  then  3 
is  tb    logarithm  of  the  number  8,  because  2^^  =  8. 

If  the  base  be  5,  then  3  is  the  logarithm  of  125,  because 
53  =  125.  ■•■•''     '  '    ^ 

63.  Properties  of  Logarithms. — The  use  of  logarithms 
depends  on  the  following  j^^'operties  which  are  true  for  all 
logarithms,  whatever  may  be  the  base. 

♦From  the  definition  it  follows  that  (1)  logaa*=x,  and  conversely  (2)  a'°Ko •»  =  »». 
Taking  the  logarithms  of  both  sides  of  the  equation  «*  =  m,  we  have  log^  a'  =  ar  =  log  to. 
Conversely,  taking  the  exponentials  of  both  sides  of  x  =  log„  m  to  base  a,  we  have 
a*  =  o'°!?o "» =  TO.  a'  =  m  and  a;  =  log„TO  are  thus  seen  to  be  equivalent,  and  to 
express  the  same  relation  between  a  number,  m,  and  its  logarithm,  :c,  to  base  a. 


■'  f 


88 


PLANE  TRIGONOMETRY. 


■  "     ''r     I 


A 


(1)  The  logarithm  of  1  is  zero. 

For  a"  =  1,  whatever  a  may  be ;  therefore  log  1=0. 

(2)  The  logarithm  of  the  base  of  any  system  is  unity.     , 
For  a'=  a,  whatever  a  may  be ;  tlierefore  log„a  =  1.       ■, 

(.3)    TJie  logarithm  of  zero  in  any  system  ivhose  base     is 
greater  than  1  is  minus  infinity, 

11 

For  a~*  =  —  =      =  0 ;  therefore  log  0  =  —  oo . 

a*      00 

(4)    The  logarithm  of  a  product  is  equal  to  the  sum  of  the 
loganthms  of  its  factors. 

For  let  a;  =  log„m,  and  .v=log„w. 

and  n  =  a'. 


r  -I 


.'.  m  =  a, 

.,  ,,  .*.  m?i  =  a'+''.  , 

.-.  Iog„wn  =  a;  +  y  =  log„m4-log„w.  ' 

Similarly,  log„mrjp  =  log„m  +  log„n  +  log„j9, 

and  so  on  for  any  number  of  factors.  '      i    .  , 

Thus,  Iog60  =  log(3x4x5),      .!    7l;.^ 

■  •         =log3  +  log4-j-log5.  ' 

(5)    The  logarithm  of  a  quotient  is  equal  to  the  logarithm 
of  the  dividend  minus  the  logarithm  of  the  divisor.  '  > 


For  let 


x  =  log„m,  and  y  =  \ogan. 


m  =  a' 


and  n  =  a*. 


m 
n 


m 


=  a 


x-y 


log«  -  =  x  —  y=i  log„  m  —  log„  w. 

17 

log— =  log  17— log  5. 
o 


PROPERTIES   OF  LOGARITHMS, 


89 


(6)  The  logarithm  of  any  poiver  of  a  number  is  equal  to 
the  logarithm  of  the  number  iriultiplied  by  the  exponent  of  the 
poiver. 

For  let  fl;  =  log.,m.     .-.  m  —  a*. 

.'.  m>'=a''*. 
.'.  log^m'' =  px  =  plog^m. 

(7)  The  logarithm  of  any  root  of  a  mimber  is  equal  to  the 
logarithm  of  the  number  divided  by  the  index  of  the  root. 

For  let  a;=log„?)i.     .-.  m  =  a'. 

1  X 

^        X      1 
.'.  log(m'-)  =  -  =  - logaWi. 
r      r 

It  follows  from  these  propositions  that  by  means  of 
logarithms,  the  operations  of  midtipUcation  and  division  are 
changed  into  those  of  addition  and  siibt r action ;  and  the 
operations  of  involution  and  evolution  are  changed  into  those 
oi  niultiplicationimCi  division. 

1.  Suppose,  for  instance,  it  is  required  to  thid  the  product 
of  246  and  357;  we  add  the  logarithms  of  the  factors,  and 
the  sum  is  the  logarithm  of  the  product :  thus, 


■  log,o246  =  2.39093 

logio357  =  2.55267 

4.94360 

which  is  the  logarithm  of  87822,  the  product  required. 

2.    If  we  are  required  to  divide  371.49  by  52.376,  we  pro- 
ceed thus:     ' 

log,o371.49=:  2.56995 

:      logio52.376  =  1.71913 

[.■,_:::■:■[  0.85082    ■■;■;.;       .  ,„,:    ■  r'. 

which  is  the  logarithm  of  7.092752,  the  quotient  required. 


"  $ 


:«  \1 


-!       .,1 


90 


PLANE  TRIGONOMETRY. 


3.   If  we  have  to  find  the  fourth  power  of  18,  we  proceed 

thus: 

logiol3=  1.11394 


■   ■      4.45576 
which  is  the  logarithm  of  285C1,  the  number  required. 

4.  If  we  are  to  find  the  fifth  root  of  1G807,  we  proceed 

thus : 

5)4.22549  =  logi„  16807, 

0.845098 
which  is  the  logarithm  of  7,  the  root  required. 

5.  Given  logio2  =  0.30103 ;  find  logi„128,  log,,, 512. 

Ans.  2.10721,  2.70927. 

6.  Given  logio3  =  0.47712 ;  find  logi„81,  logi„2187. 

Ans.  1.90849,  3.33985. 

_   7.    Given  log,o3;  find  log,,,  W'.  ...    ,     ,      0.28627. 

8.   Find  the  logarithms  to  the  base  a  of 


10 


v/a.   Vet-. 


9.   Find  the  logarithms  to  the  base  2  of  8,  04,  ^,  .125, 


.015625,  V64. 


Ans.  3,  6,   —  1, 


3, 


6,  2. 


10.    Find   the   logarithms   to  base   4   of    8,    V 16,    V.5> 
^.015625.  Ans.  |,  |,   -  \,   -  1. 

Express  the  following  logarithms  in  terms  of  log  a,  log?>, 
and  logo:  ,  ,    ,. 

Ans.  6  log  a -f- 0  log  6  4- 3  log  c. 


11.  log -/ {on'' cy. 

12.    log^a^^'FZ. 


13.    log 


t'   f'St 


• 

(a-n-'c-*)^ 


floga+flog6  +  |log, 
iloga. 


SYSTEMS   OF  LOGARITHMS. 


91 


64.  Common  System  of  Logarithms.  —  There  are  two 
systems  of  logarithms  in  use,  viz.,  the  Naperian*  system 
and  the  common  system.  '  .  . 

The  Naperian  system  is  used  for  purely  theoretic  investi- 
gations; its  base  is  e  =  2.7182818. 

•  The  common  system  f  of  logarithms  is  the  system  that 
is  used  in  all  practical  calculations ;  its  base  is  10. 

By  a  system  of  logarithms  to  the  base  10,  is  meant  a  suc- 
cession of  values  of  x  which  satisfy  the  equation 

■  -.  .."   ;     •.  -7)1  =  10', 

for  all  positive  values  of  m,  integral  or  fractional.  Thus,  if 
Ave  suppose  m  to  assume  in  STiccession  every  value  from  0 
to  00,  the  corresponding  values  of  x  will  form  a  system  of 
logarithms,  to  the  base  10.  . 

Such  a  system  is  fo  -med  by  means  of  the  series  of  loga- 
rithms of  the  natural  numbers  from  1  to  100000,  which  con- 
stitute the  logarithms  registered  in  our  ordinary  tables. 


J/'* 


Now 


and  so  on. 
Also, 


10"  =1, 
10^  =  10, 
10'  =  100, 
lO''  =  1000, 


logl  =0; 
log  10  =1; 
log  100  =2; 
log  1000  =  3. 


10-'=, V      =•!.         •••  log.l       =-1; 
_      ,     .        10-^=  ik    =-01.       •••  log-01     =-2; 

■' " ;    ■      io-»=  i-^o  = -^^1?    •*•  log.OOl  =-3. 
and  so  on. 

Hence,   in   the   common   system,  the  logarithm   of  any 

number  between 

1  and  10  is  some  number  between  0  and  1 ;    i.e.,  0  -}- 
a  decimal ; 

*  So  cnllerl  from  its  inventor,  7?«»'on  iV«/>i>/*,  a  Scotch  motliematlcian. 
t  First  introduced  in  lOl;')  by  liriygs,  a  contemporary  of  Napier. 


t 


1 


t! 

I 


92 


PLANE  TRIGONOMETBY. 


IV. 


' 


10  and  100  is  some  number  between  1  and  2 ;  i.e.,  1  + 
a  decimal;  .  -^  ;i 

100  and  1000  is  some  number  between  2  and  3 ;  i.e.,  2  -f 
a  decimal;  "    •  ^ 

1  and  .1  is  some  number  between  0  and  —  1 ;  i.e.,  —  1  -f- 
a  decimal ; 

.1  and  .01  is  some  number  between   —  1  and  —  2 ;  i.e., 

—  2  +  a  decimal ; 

.01  and  .001  is  some  number  between   —  2  and  —  3 ;  i.e., 

—  3  +  a  decimal ; 

and  so  on. 

It  thus  appears  that 

(1)  The  (common)  logarithm  of  any  number  greater  than 
1  is  positive. 

(2)  The  logarithm  of  any  positive  number  less  than  1  is 
negative. 

(3)  In  general,  the  common  logarithm  of  a  number  con- 
sists of  two  parts,  an  integral  part  and  a  decimal  part. 

The  integral  part  of  a  logarithm  is  called  the  characteristic 
of  the  logarithm,  and  may  be  either  positive  or  negative. 

The  decimal  part  of  a  logarithm  is  called  the  mantissa 
of  the  logarithm,  and  is  ahcays  kept  2wsitive. 

Note.  — It  is  convenient  to  keep  the  decimal  part  of  the  lognrithma  alwaya  posi- 
tive, in  order  tliat  numbers  couBisting  of  the  same  digits  in  the  same  order  may 
correspond  to  the  same  mantissa. 

It  is  evident  from  the  above  examples  that  the  character- 
istic of  a  logarithm  can  always  be  obtained  by  the  following 
rule : 

Rule.  —  The  characteristic  of  the  logarithm  of  a  number 
greater  than  unity  is  one  less  than  the  number  of  digits  in 
the  whole  number. 

The  characteristic  of  the  logarithm  of  a  number  less  than 
unity  is  negative,  and  is  one  more  than  the  number  of  ciphers 
immediately  after  the  decimal  point. 


RULES  FOR   THE  CHARACTERISTIC. 


93 


Thus,  the  characteristics  of  the  logarithms  of  1234,  123.4, 
1.234,  .1234,  .00001234, 12340,  are  respectively,  3,  2,  0,  - 1, 

-6,  4.      ^^  ■     •'"  '        ■    ' 

Note.  —  When  the  characteristic  Is  uegative,  the  minus  sign  is  written  over  it  to 
indicate  that  the  characteristic  alone  is  negative,  the  mantissa  being  always  positive. 

Write  down  the  characteristics  of  the  common  logarithms 
of  the  following  numbers  : 

1.  17601,  361.1,  4.01,  723000,  29.  Ans.  4,  2,  0,  5,  1. 

2.  .04,  .0000612,  .7963,  .001201,  .1. 

Ans.  -2,  -5,  -1,  -3,  -1. 

3.  How  many  digits  are  there  in  the  integral  part  of  the 
numbers  whose  common  logarithms  are  respectively  3.461, 
0.30203,  5.47123,  2.67101  ? 

4.  Given  log  2  =  0.30103 ;  find  the  number  of  digits  in  the 
integral  part  of  8^",  2'^,  16"^,  2'''.  Ans.  10,  4,  25,  31. 

65.  Comparison  of  Two  Systems  of  Logarithms.  —  Given 
the  logarithm  of  a  number  to  base  a ;  to  find  the  logarithm 
of  the  same  number  to  base  b. 

Let  m  be  any  number  whose  logarithm  to  base  b  is 
required. 

Lfll  x  =  \ogf,m;  then  b'  =  m. 


V^ 


loga(60  =  log„m;  or  a;log„6  =  log„m. 
1 


.*.   05  = 


log„6 


X  log<.m. 


or 


logsm  = 


log„6 


(1) 


•Hence,  to  transform  the  logarithm  of  a  number  from 

1 


base  a  to  base  b,  we  multiply  it  by 


loga^ 


94 


PLANE  TRIGONOMETRY. 


This  constant  multiplier 


log„6 


is  called  the  modulus  of 


tl    i 


I?  f 


the  system  of  wJiich  the  base  is  b  with  reference  to  the  system 
of  which  the  base  is  a. 

If,  then,  a  list  of  logarithms  to  some  base  e  can  be  made, 
we  can  deduce  from  it  a  list  of  common  logarithms  by  mul- 
tiplying each  logarithm  in  the  given  list  by  the  modulus 

of  the  common  system 

Putting  a  for  m  in  (1),  we  have 

c.      r  . :       log»a  =  J^  =  r^,  by  (2)  of  Art.  63. 

log„6      log„6 

r      ;    ,-.    logja  X  log„&=l.  ; 


,,„^     ^,,  ;,„^,^, ,,,.;,   ,,^,,^,,     EXAMPLES,     ■'    ,»..■,      - 

1.  Show  how  to  transform  logarithms  with  base  5  to 
logarithms  with  base  125. 

Let  m  be  any  number,  and  let  x  be  its  logarithm  to  base 
125. 
Then  m  =  125'  =  (5»)'  =  5^^     .-.  3a;  =  \og,m. 

.-.  a;  =  logi25m  =  |log5m.  ;^    • 

Thus,  the  logarithm  of  any  number  to  base  5,  divided  by 
3  (i.e.,  by  logs  125),  is  the  logarithm  of  the  same  number  to 
the  base  125. 

Otherwise  by  the  rule  given  in  (1) .     Thus, 

log5l25         o 
Show  how  to  transform 

2.  Logarithms  with  base  2  to  logarithms  with  base  8. 

Ans.  Divide  each  logarithm  by  3. 


iti 


TABLES  OF  LOGARITHMS. 


95 


iW 


3.  Logarithms  with  base  9  to  logarithms  with  base  3. 

Ans.  Multiply  each  logarithm  by  2. 

4.  Find  logs  8,  logjl,  log8  2,  logyl,  log32l28. 

Ans.  3,  0,  i,  0,  i. 

66.  Tables  of  Logarithms.  —  The  common  logarithms  of 
all  integers  from  1  to  100000  have  been  found  and  registered 
in  tables,  which  are  therefore  called  tabular  logarithms.  In 
most  tables  they  are  given  to  six  places  of  decimals,  though 
they  may  be  calculated  to  various  degrees  of  approximation, 
such  as  five,  six,  seven,  or  a  higher  number  of  decimal  places. 
Tables  of  logarithms  to  seven  places  of  decimals  are  in 
common  use  for  astronomical  and  mathematical  calculations. 
The  common  system  to  base  10  is  the  one  in  practical  use, 
and  it  has  two  great  advantages : 

(1)  From  the  rule  (Art.  64)  the  characteristics  can  be 
written  down  at  once,  so  that  only  the  mantissse  have  to  be 
given  in  the  tables. 

(2)  The  mantissse  are  the  same  for  the  logarithms  of  all 
numbers  which  have  the  same  significant  digits,  in  the  same 
order,  so  that  it  is  sufiicient  to  tabulate  the  mantissse  of  the 
logarithms  of  integers. 

For,  since  altering  the  position  of  the  decimal  point  with- 
out changing  the  sequence  of  figures  merely  multiplies  or 
divides  the  number  by  an  integral  power  of  10,  it  follows 
that  its  logarithm  will  be  increased  or  diminished  by  an 
integer;  i.e.,  that  the  mantissa  of  the  logarithm  remains 
unaltered. 

In  General.  —  If  N  be  any  number,  and  p  and  q  any 
integers,  it  follows  that  N  x  10^  and  N  -f- 10'  are  numbers 
whose  significant  digits  are  the  same  as  those  of  N. 

Then    log  (N  x  10^)  =  log N  4- i> log  10  =  logN  -{-p.     (1) 
Also,     log  (N  -5- 10')  =  logN  -  gloglO  =  logN  -  q.     (2) 


96 


PLANE  TRIGONOMETRY. 


!■■     .  ( 


ar* 


In  (1)  the  logarithm  of  N  is  increased  by  an  integer,  and 
in  (2)  it  is  diminished  by  an  integer. 

That  is,  the  same  mantissa  serves  for  the  logarithms  of 
all  numbers,  whether  greater  or  less  than  unity,  which  have 
the  same  significant  digits,  and  differ  only  in  the  position 
of  the  decimal  point. 

This  will  perhaps  be  better  understood  if  we  take  a 
particular  case. 

From  a  table  of  logarithms  we  find  the  mantissa  of  the 
logarithm  of  787  to  be  895975 ;  therefore,  prefixing  the  char- 
acteristic with  its  appropriate  sign  according  to  the  rule, 
we  have  ,       ,      ,  .    . 

log787     =2.895975.  .         ,    ,       :    /      . 

Now  log  7.87    =  log  ^  =  log  787 -2 

=  0.895975.  , 
Also,  log.0787  =log(j55^)^    ^og787-4     , 

'   ' .    -  J':  .  .  .  =2.895975.  '.';.  ,  '.r  :    .  ,.:  ■. 

Also,  log  78700  =  log  (787  x  100)  =  log  787  +  2 

■  r.'.-'  <-:.,:    '        =4.895975.    '^■^-''  :    . '■         ' 


Note  1.  —  We  do  not  write  log,o  787 ;  for  so  long  as  we  are  treating  of  logarithms 
to  the  particular  base  10,  we  may  omit  the  suffix. 

Note  2.  —  Sometimes  in  woriiing  with  negative  logarithms,  an  arithmetic  artifice 
will  be  necessary  to  make  the  mantissa  positive.  For  example,  a  result  such  as 
—  2.69897,  in  which  the  whole  expression  is  negative,  may  be  transformed  by  sub- 
tracting 1  from  the  characteristic,  and  adding  1  to  the  mantissa,    Thus, 

. ,  -  2.69897  =  -  3  +  (1  -  .69897)  =  3.30103. 

NoTB  3.  —  When  the  characteristic  of  a  logarithm  is  negative,  it  is  often,  espe- 
cially in  Astronomy  and  Geodesy,  for  convenience,  made  positive  by  the  addition 
of  10,  which  can  lead  to  no  error,  if  we  are  careful  to  subtract  10. 

Thus,  instead  of  the  logarithm  3.608582,  we  may  write  7.603582  —  10. 

In  calculations  with  negative  characteristics  we  follow 
the  rules  of  Algebra. 


EXAMPLES. 


97 


1.   Add  together 


EXAMPLES. 

2.2143 

.       1.3142 

6.9068 


2.    From 
take 


7.4353  Ans. 

3.24569 
5.62493 


1.62076 


the  1  carried  from  the  last  subtraction  in  decimal  places 
changes  —  5  into  —  4,  and  then  —  4  subtracted  from  —  3 
gives  1  as  a  result, 

3.  Multiply  2.1528  by  7. 

2.1528  • 

7 

.,  13.0696  .      .  :     ■ 

the  1  carried  from  the  last  multiplication  of  the  decimal 
places  being  added  to  —  14,  and  thus  giving  —  13  as  a  result. 

Note  4.  —  When  a  logarithm  with  negative  characterietic  has  to  be  divided  by  a 
number  which  ie  not  an  exact  divisor  of  the  characteristic,  we  proceed  as  follows  in 
order  to  keep  the  characteristic  integral.  Increase  the  characteristic  numerically  by 
a  number  which  will  make  it  exactly  divisible,  and  prefix  an  equal  positive  number 
to  the  mantissa. 

4.  Divide  3.7268  by  5. 

Increase  the  negative  characteristic  so  that  it  may  be 
exactly  divisible  by  5 ;  thus  - 


3.7268 


^  5  -f  2.7268  ^  j- 


5  5 


5453. 


Given  that  log2=.30103,  log3=:.47712,  and  log7=.84510; 
find  the  values  of 

5.   log6,  log42,  logl6.  ^ri5.  .77815,  1.62325,  1.20412. 


«  ll' 


98 


PLANE  TRIGONOMETRY. 


i 


n  fi 


Ipifj 


6.  log 49,  log 36,  log 63.     Ans.  1.69020,  1.55630,  1.79934. 

7.  log 200,  log  600,  log  70.  2.30103,2.77815,1.84510. 

8.  log 60,  log. 03,  log  1.05,  log. 0000432. 

Note.  —  The  logarithm  of  5  and  its  powers  can  always  be  obtained  from  log 2. 

Ans.  1.77815,  2.47712,  .02119,  5.63548. 

9.  Given  log  2 = .30103 ;  find  log  128,  log  125,  and  log  2500. 

Ans.  2.10721,  2.09691,  3.39794. 

Given  the  logarithms  of  2,  3,  and  7,  as  above ;  find  the 
logarithms  of  the  following : 

10.  20736,  432,  98,  686,  1.728,  .336. 

Ans.  4.31672,  2.63548,  1.99122,  2.83632,  .23754,1.52634. 

11.  V.2,  (.03) ^  (.0021)^,  (.098)^  (.00042)^  (.0336)*. 
Ans.  1.65052,  1.61928, 1.46444,  4.97368,  17.11625, 1.26317. 

67.  Use  of  Tables  of  Logarithms  *  of  Numbers.  —  In  our 

explanations  of  the  use  of  tables  of  common  logarithms  Ave 
shall  use  tables  of  seven  places  of  decimals.f  These  tables 
are  arranged  so  as  to  give  the  mantissie  of  the  logarithms 
of  the  natural  members  from  1  to  100000 ;  i.e.,  of  numbers 
containing  from  one  to  five  digits. 

A  table  of  logarithms  of  numbers  correct  to  seven  deci- 
mal places  is  exact  for  all  the  practical  purposes  of  Astron- 
omy and  Geodesy.  For  an  actual  measurement  of  any  kind 
must  be  made  with  the  greatest  care,  with  the  most  accurate 
instruments,  by  the  most  skilful  observers,  if  it  is  to  attain 
to  anything  like  the  accuracy  represented  by  seven  signifi- 
cant figures. 

*  The  methods  by  which  these  tables  are  formed  will  be  given  in  Chap.  VIII. 

t  The  student  should  here  provide  himself  with  logarithmic  and  trigonometric 
tables  of  seven  decimal  places.  The  most  convenient  seven-flgure  tables  used  in  this 
country  are  Stanley's,  Vega's,  Bruhns',  etc.  In  the  appendix  to  the  Elementary 
Trigonometry  are  given  five-flgurod  tables,  which  are  sufficiently  near  for  most  prac- 
tical application!. 


,Jt 


USE  OF  LOGARITHMIC   TABLES. 


99 


If  the  measure  of  any  length  is  known  accurately  to  seven  figures, 
it  is  practically  exact ;  i.e.,  it  is  known  to  within  the  limittt  of  obser- 
vation. 

If  the  measure  of  any  angle  is  known  to  within  the  tenth  part  of  a 
second,  the  greatest  accuracy  possible,  at  present,  in  the  measurement 
of  angles  is  reached.  The  tenth  part  of  a  second  is  about  the  two- 
millionth  part  of  a  radian.  This  degree  of  accuracy  is  attainable  only 
with  the  lai'gest  and  best  instruments,  and  under  the  most  favorable 
conditions. 

On  page  101  is  a  specimen  page  of  Logarithmic  Tables. 
It  consists  of  the  mantissse  of  the  logarithms,  correct  to 
seven  places  of  decimals,  of  all  numbers  between  62500  and 
63009.  The  figures  of  the  ninnher  are  those  in  the  left 
column  headed  N,  followed  l)y  one  in  larger  type  at  the  top 
of  the  page.  The  first  three  figures  of  the  mantisste  795, 
796,  797  •••,  and  the  remaining  four  are  in  the  same  hori- 
zontal line  with  the  first  four  figures  of  the  number,  and  in 
the  vertical  column  under  the  last. 

Logarithms  are  in  general  incommensurable  numbers. 
Their  values  can  therefore  only  be  given  approximately. 
Throughout  all  approximate  calculations  it  is  usual  to  take 
for  the  last  figure  whicli  we  retain,  that  figure  which  gives 
the  nearest  aj^proach  to  the  true  value.  When  only  a  cer- 
tain number  of  decimal  places  is  required,  the  general  rule 
is  this:  Strike  out  the  rest  of  the  Jirjures,  and  increase  the  last 
figure  retained  by  1  if  the  first  figure  struck  off  is  5  or  greater 
than  5. 

68.  To  find  the  Logarithm  of  a  Given  Number.  —  When 
the  given  number  has  not  more  than  five  digits,  we  have 
merely  to  take  the  mantissa  immediately  from  the  table, 
and  prefix  the  characteristic  by  the  rule  (Art.  64). 

Thus,  suppose  Ave  require  the  logarithm  of  62541.  The 
table  gives  .7961648  as  the  mantissa,  and  the  characteristic 
is  4,  by  the  rule ;  therefore 

log    62541  =  4.7961648. 
Similarly,  log  .006281  =  3.7980288     .     .     (Art.  64) 


100 


PLANE  TRIGONOMETRY. 


WM 


My 


t 


1 


Suppose,  however,  that  the  given  number  has  more  than 
five  digits.     For  example : 

Suppose  we  require  to  find  log  627G1.6. 
We  find  froxn  the  table 

log62761  =  4.7976899 
log  62762  =  4.7976968 


and 


diff.  for  1  =  0.0000069 


Thus  for  an  intrease  of  1  in  the  number  there  is  an  in- 
crease of  .0000069  in  the  logarithm. 

Hence,  assuming  that  the  increase  of  the  logarithm  is 
proportional  to  the  increase  of  the  number,  then  an  increase 
in  the  number  of  .6  will  correspond  to  a.i  increase  in  the 
logarithm  of  ,6  x  .0000069  =  .0000041,  to  the  nearest  sev- 
enth decimal  place. 


Hence, 


log  62761  =  4.7976899 
diff.  for  .6  =  41 


log  62761.6  =  4.7976940 


This  explains  the  use  of  the  column  of  proportional  parts 
on  the  extreme  right  of  the  page.  It  will  be  seen  that  the 
difference  between  the  logarithms  of  two  consecutive  num- 
bers is  not  always  the  same ;  for  instance,  those  in  the 
upper  part  of  the  page  before  us  differ  by  .0000070,  while 
those  in  the  middle  and  the  lower  parts  differ  by  .0000069 
and  .0000068.  Under  the  column  with  the  heading  69  we 
see  the  difference  41  corresponding  to  the  figure  6,  which 
implies  that  when  the  difference  between  the  logarithms  of 
two  consecutive  members  is  .0000069,  the  increase  in  the 
logarithm  corresponding  to  an  increase  of  .6  in  the  number 
is  .0000041 ;  for  .06  it  is  evidently  .0000004,  and  so  on. 

Note.  —  We  assume  in  Uiis  method  that  the  increase  in  a  logarithm  is  propor- 
tional to  the  increase  in  the  number.  Although  this  is  not  strictly  true,  yet  it  is  in 
most  cages  sufllciently  exact  for  practical  purposes. 

Had  we  talven  a  whole  number  or  a  decimal,  the  process  would  have  been  the 
■ame. 


1* 


TABLE  OF  LOOARITHMS. 


101 


N. 

6250 

51 
52 
58 
54 

55 
56 
57 

58 

59 

6260 

61 
62 
63 
64 

65 
66 
67 
68 
69 

6270 

71 
72 
73 
74 

75 
76 

77 
78 
79 

6280 

81 
82 
83 
84 

85 
86 
87 
88 
89 

6290 
91 
92 
93 
94 

95 
96 
97 

98 
99 

6800 

N. 


795  8800 


9495 

796  0190 

0884 

1579 

2273 
2%7 
3662 
4356 
5050 


8870 

9564 
0259 
0954 
1648 

2343 
3037 
3731 


4425  4494 
5119  5188 


796  5743 

6437 
7131 
7824 
8517 
9211 
9904 
797  0597 
1290 
1983 


797  2675 


6506 
7200 
7893 
8587 

9280 
9973 
0666 
1359 
2052 

2745 


3368 
4060 
4753 
5445 

6137 
6829 
7521 
8213 
8905 


797  9596 


798  0288 
0979 
1671 
2362 

3053 
3744 
4435 
5125 
5816 


798  6506 


7197 
7887 
8577 
9267 

9957 
799  0647 
1337 
2027 
2716 


799  3405 
O 


8939 


9634 
0329 
1023 
1718 

2412 
3106 
3800 


5813 


3437 
4130 
4822 
5514 

6207 
6899 
7590 
8282 
8974 


9666 

0357 
1048 
1740 
2431 

3122 
3813 
4504 
5194 

5885 


6575 


7266 
7956 
8646 
9336 

0026 
0716 
1406 
2096 
2785 


3474 
1 


5882 

6576 
7269 
7963 
8656 

9349 
0043 
0736 
1428 
2121 

2814 

3507 
4199 
4891 
5584 

6276 
6968 
7660 
8351 
9043 


5951 

6645 
7339 
8032 
8725 

9419 
0112 
0805 
1498 
2191 


97^ 
0426 
1118 
1809 
Z500 

3191 

3882 
4573 
5263 
5954 


6645 


7335 
8025 
8715 
9405 

0095 
0785 
1475 
2164 
2854 


9009 

9703 
0398 
1093 

1787 

2481 
3176 
3870 
4564 

5258 


9078 

9773 
0468 
1162 
1857 

2551 
3245 
3939 
4633 
532^ 

6021 


2883 


3576 
4268 
4961 
5653 

6345 
7037 
7729 
8421 


6714 

7408 
8101 
8795 
9488 
0181 
0874 
1567 
2^60 

2952 

3645 
4337 
5030 

5722 

6414 
7106 
7798 
8490 
911219181 


9804  9873 


0495 
1187 
1878 
2569 

3260 
3951 
4642 
5333 
6023 


6714 


3543 


7404 
8094 
8784 
9474 

0164 
0854 
1544 
2233 


0565 
1256 
1947 
2638 

3329 
4020 
4711 
5402 
6092 


6783 


7473 
8163 
8853 
9543 

0233 
0923 
1613 
2302 


2923  2992 


3612 


3681 


9148 


9842 
0537 
1232 
1926 
2620 
3314 
4009 
4703 
5396 

6090 
6784 
7477 
8171 
8864 


6 

9217 

9912 
0606 
1301 
1995 
2690 
3384 
4078 
4772 
5466 

6160 

6853 
7547 
8240 
8933 


9557  9627 
0250  0320 


0943 
1636 
2329 

3022 


3714 
4407 
5099 
5  791 
6483 
7175 
7867 
8559 
9251 


9942 


0634 
1325 
2016 
2707 

3398 
4089 
4780 
5471 
6161 

6852 

7542 
8232 
8922 
9612 

0302 
0992 
1682 
2371 
3061 


3750 


1013 
1706 
2398 

3091 

3784 
4476 
5168 
5860 

6553 
7245 
7936 
8628 
9320 


0011 


0703 
1394 

2085 
2776 

3467 
4158 
4849 
5540 
6230 


6921 

7611 
8301 
8991 
9681 

0371 
1061 
1751 
2440 
3130 


3819 
6 


9m 

9981 
0676 
1370 
2065 

2759 
3453 
4147 
4841 
5535 

6229 

6923 
7616 
8309 
9003 

9696 
0389 
1082 
1775 
2468 

3160 

3853 
4545 
5237 
5930 

6622 
7314 
8006 
8697 
9389 


0080 


0772 
1463 
2154 
2846 

3536 
4227 
4918 
5609 
6299 

6990 


7680 
8370 
9060 
9750 

0440 
1130 
1820 
2509 
3199 
3888 


8   9 


9356  9426 


0051,0120 


0745 
1440 
2134 


0815 
1509 
2204 


28292898 
35233592 
•12174286 
49114980 
5605:5674 


629816368 

6992  7061 

7685  7755 
8379  8448 


9072 


9141 


9765  9835 
0458  0528 


1151 
1844 
2537 
3229 
3922 
4614 
5307 
5999 

6691 
7383 
8075 
8766 
9458 


0150 


0841 
1532 
2224 
2915 

3606 
4296 
4987 
5678 
6368 

7059 


7749 
8439 
9129 
9819 

0509 
1199 
1889 
2578 
3268 


1221 
1913 
2606 

3299 

3991 
4684 
5376 
6068 

6760 
7452 
8144 
8836 
9527 

0219 


0910 
1601 
2293 
2984 

3675 
4366 
5056 
5747 
6437 


7128 


3957 


7818 
8508 
9198 
9888 

0578 
1268 
1958 
2647 
3337 


4026 
9 


P.P. 


70 

7.0 

14.0 

21.0 

28.0 

35.0 
42.0 
49.0 
56.0 

63.0 


69 
6.9 
13.8 
20.7 
27.6 
34.5 
41.4 
48.3 
55.2 


9  62.1 


68 

6.8 

13.6 

20.4 

27.2 

34.0 
40.8 
47.6 
54.4 

61.2 


i 


*. 


Sf:# 


p.p. 


\m 


102 


PLANE  TRIGONOMETRY. 


> 


Tbua,  Huppotc  we  require  to  And  log  0-27U10  uiul  lug  .627010.  TIil-  inanllijiiu  In 
exactly  the  auine  us  before  (Art.  00),  and  the  only  dltt'urenue  tu  be  iiiudc  io  the  tlual 
reault  it  to  cbauge  tbe  ciiaracteriatiu  uccurding  to  rule  (Art.  04). 

Tbu»  log  027010  =  u.7U7e942, 

and  log  .627016  =  1.7970942. 

69.  To  find  the  Number  corresponding  to  a  Oiven  Loga- 
rithm. —  It'  the  decimal  part  of  tlie  logarithm  is  found  ex- 
actly ill  the  table,  we  can  take  out  the  corresponding 
number,  and  put  the  decimal  point  in  the  number,  in  the 
place  indicated  by  the  characteristic. 

Thus  if  we  have  to  tind  the  number  whose  logarithm  is 
2.7982915,  we  look  in  the  table  for  the  mantissa  .7982915, 
and  we  find  it  set  down  opposite  the  number  62848 :  and 
as  the  characteristic  is  2,  there  must  be  one  cipher  before 
the  first  significant  figure  (Art.  G4). 

Hence  2.7982915  is  the  logarithm  of  .062848. 

Next,  suppose  that  the  decimal  part  of  the  logarithm  is 
not  found  exactly  in  the  table.  For  example,  suppose  we 
have  to  find  the  number  whose  logarithm  is  2.7974453. 

We  find  from  the  table 

log62726  =  4.7974476 
log62725  =  4.7974407 

diff .  for  1  =  ".0000069 

Thus  for  a  difference  of  1  in  the  numbers  there  is  a 
ditt'erence  of  .0000069  in  the  logarithms.  The  excess  of 
the  given  manl.ssa  above  .7974407  is  (.7974453  -  .7974407) 
or  .0000046. 

Hence,  assuming  that  the  increase  of  the  number  is 
proportional  to  the  increase  of  the  logarithm,  we  have 

.0000069  :  .0000046  :  :  1 :  number  to  be  added  to  627.25. 

...  number  to  be  added  =  -^^-^^^'  =  1^'=. 667  fi9)46.0(.666 

.0000069      69  41 4 

.-.  log62725.667  =  4.7974453, 

and      .-.  log627.25667  =  2.7974453; 

therefore  number  required  is  627.25667. 


4  60 
414 

~460 


ARITHMETIC  COMPLEMENT. 


108 


We  might  have  saved  the  labor  of  dividing  40  by  09,  by 
using  the  table  of  proportional  parts  as  follows : 


given  mantissa      =  .7074453 
mantissa  of  02725  =  .7974407 

diff.  of  mantissie  =  40 

proportional  part  for  .0      =  41.4 


r- .'  I' '. 


(( 


it 


''•' 

4.6 

n 

(C 

.00   = 

4.14 
.46 

« 

(( 

.000  = 

.414 

.  .           and  so  on 

number  = 

027.250000..  •. 

1 


69a.  Arithmetic  Complement.  —  By  the  arithmetic  com- 
plement of  the  logarithm  of  a  number,  or,  briefly,  the 
cologarithm  of  the  number,  is  meant  the  remainder  found 
by  subtracting  the  logarithm  from  10.  To  subtract  one 
logarithm  from  another  is  the  same  as  to  add  the  co- 
logarithm  and  then  subtract  10  from  the  result.  ,    .     .. 


Thus, 


a 


6  =  a  -4-  (10  -  6)  -  10, 


where  a  and  b  are  logarithms,  and  10  —  6  is  the  arithmetic 
complement  of  6. 

When  one  logarithm  is  to  be  subtracted  from  the  sum  of 
several  others,  it  is  more  convenient  to  add  its  cologarithm 
to  the  sum,  and  reject  10.  The  advantage  of  using  the 
cologarithm  is  that  it  enables  us  to  exhibit  the  work  in  a 
more  compact  form. 

The  cologarithm  is  easily  taken  from  thr  table  mentally 
by  subtracting  the  last  significant  figure  on  the  right  from 
10,  and  all  the  others  from  9. 


104 


PLANE  TRIGONOMETRY. 


1.   Given 

find 
1                     2.   Given 

wi 

1                  find 

1                      3.   Given 

m               find 

f                    4.   Given    • 

find           - 

6.  Given 

find  the  number 

6.   Given 

.1    ':•' 

find  the  number 

7.   Given 

find  the  number 

8.   Given      . 

I 

find  the  number 

Ans.  4.7201799. 


EXAMPLES. 

log  52502  =4.7201758, 
log  52503  =4.7201841; 
log  52502.5. 

log  3.0042  =  0.4777288, 

log  3.0043  =  0.4777433;  J 

log  300.425.  ■    '  2.4777360. 

■  log  7.6543  =  0.8839055, 

■  log  7.6544  =  0.8839112; 

log  7.65432.  .8839066. 

log  6.4371  =  0.8086903, 

log  6.4372  =  0.8086970; 

log  6437125.  6.8086920. 

log  12954  =4.1124039, 
log  12955  =4.1124374; 
whose  logarithm  is  4.1124307.  12954.8. 

log  60195  =  4.7795532,      '  - 
log  60196  =4.7795604; 
whose  logarithm  is  2.7795561.      601.95403. 

'  log3.7040  =  .5686710,  ^  ' '   ' 

log  3.7041  =  .5686827 ;  ,      ;-  ,   , 

whose  logarithm  is  .5686760.  3.70404. 

log  2.4928  =  .3966874,. 
log  2.4929  =  .3967049 ; 
whose  logarithm  is  6.3966938.         2492837. 


NATURAL   TRIGONOMETRIC  FUNCTIONS. 


105 


find 


9.   Given  log  32642  ==  4.5137768, 

log  32643  =-.4.5137901; 
log  32642.6.  '' 


10.   Find  the  logarithm  of  62654326. 
Use  specimen  page. 


Ans.  4.5137835. 
7.7969510. 


11.   Find  the  number  whose  logarithm  is  4.7989672. 

Ans.  62945.876. 

70.  Use  of  Trigonometric  Tables.  — Trigonometric  Tables 
are  of  two  kinds,  — Tables  of  Natural  Trigonometric  Functions 
and  Tables  of  Logarithmic  Trigonometric  Functions.  As  the 
greater  part  of  the  computations  of  Trigonometry  is  carried 
on  by  logarithms,  the  latter  tables  are  by  far  the  most  use- 
ful. 

We  have  explained  in  Art.  27  how  to  find  the  actual 
numerical  values  of  certain  trigonometric  functions,  exactly 
or  approximately.  '  ' 

Thus,      sin  30°  =  -;  that  is,  .5  exactly. 

Also,      tan  60°  =  V3 ;  that  is,  1.73205  approximately. 

A  table  of  natural  trigonometric  functions  gives  their 
approximate  numerical  values  for  angles  at  regular  intervals 
in  the  first  quadrant.  In  some  tables  the  angles  succeed 
each  other  at  intervals  of  1",  in  others,  at  intervals  of  10", 
but  in  ordinary  tables  at  intervals  of  1':  and  the  values  of 
the  functions  are  given  correct  to  five,  six,  and  seven  places. 
The  functions  of  intermediate  angles  can  be  found  by  the 
principle  of  proportional  parts  as  applied  in  the  table  of 
logarithms  of  numbers  (Arts.  68  and  69). 

It  is  sufficient  to  have  tables  which  give  the  functions 
of  angles  only  in  the  first  quadrant,  since  the  functions  of 
all  angles  of  whatever  size  can  be  reduced  to  functions 
of  angles  less  than  90°  (Art.  35). 


■  '■•■«. 


106 


PLANE  TRIGONOMETRY. 


'141 


71.  Use  of  Tables  of  Natural  Trigonometric  Functions.  — 

These  tables,  which  consist  of  the  actual  numerical  values 
of  the  trigonometric  functions,  are  commonly  called  tables 
of  natural  sines,  cosines,  etc.,  so  as  to  distinguish  them  from 
the  tables  of  the  logarithms  of  the  sines,  cosines,  etc. 

We  shall  now  explain,  first,  how  to  determine  the  value 
of  a  function  that  lies  between  the  functions  of  two  con- 
secutive angles  given  in  the  tables ;  and  secondly,  how  to 
determine  the  angle  to  which  a  given  ratio  corresponds. 

S' 

72.  To  find  the  Sine  of  a  Given  Angle. 

Find  the  sine  of  25°  14'  20",  having  given  from  the  table 

sin2o°lo'  =  .42GoG87       ,  ^  - 
sin  25°  14' =  .4203056        !.;    : 


d 


=  .0000877. 


diff.  for  1' =  .0002631  ' 

Let  (?  =  diff.  for  20" ;  and  assuming  that  an  increase  in 
the  angle  is  proportional  to  an  increase  in  the  sine,  we  have 

60  :  20  ::  .0002031 :  (^ 

20  X  .0002031 
60 

.-.  sin 25°  14' 20'- =.4203056 +  .0000877 
•  =.4263933.      ' 

Note.  —  We  asnumed  here  tliaf.  iin  increase  in  tlie  angle  ia  proportional  to  the 
increase  in  the  corresponding  sine,  wliich  is  suiiiciently  exact  for  practical  purposes, 
with  certain  exceptions. 

73.  To  find  the  Cosine  of  a  Given  Angle. 

Find  the  cosine  of  44°  35'  25",  having  given  from  the  table 

cos  44°  35' =  .7122303 
cos  44°  30'  =  .7120260 


dife.  for  1'  =  .0002043 

observing  that  the  cosine  decreases  as  the  angle  increases 
from  0°  to  90°. 


m 
th 

CO 


hi 


f 


EXAMPLES. 


107 


Let  d  —  decrease  >  'f  cosine  for  25"  ;  then 
GO  :  25  :  :  .0002043  :  d. 


25 


.-.  rf  =  ~  X  .0002043  =  .0000851. 

m 

.'.  cos 44°  35'  25"  =  .7122303  -  .0000851 
'■         '  =7121452. 

Similarly,  we  may  find  the  values  of  the  other  trigono- 
metric functions,  remembering  that,  in  the  lirst  quadrant, 
the  tangent  and  secant  increase  and  the  cotangent  and 
cosecant  decrease,  as  the  angle  increases. 


1. 

Given 

find 

2. 

Given 

find 

3. 

Given 

find 

4. 

Given 

find 

•y,  , 

5. 

Given 

find 

EXAMPLES. 

sin  44°  35'=  .7019459, 
sin  44°  36'=  .7021531; 
sin  44°  35'  25". 

sin42°15'=.G723GG8, 
sin  42°  1G'=  .0725821; 
sin  42°  15'  IG". 

sin  43°  23'=  .G8G87G1, 
sin43°22'=.G8GG647; 
sin  43°  22'  50". 

sin  31°  G'  =.5105333, 
sin  31°  7'  =.51G7824; 
sin  31°  G' 25". 

cos74°45'=.42G5G87, 
cos  74°  40' =.4263056; 
cos  74°  45'  40". 


Ahs.  .7020322. 


.6724242. 


.6868408. 


.5166371. 


.4263933. 


■* 


M 

1       '^^^ 

H 

m                     6.    Given 

1 

1                  ^"^ 

I 

m                     7.   Given 

-■1*  ).■  (;; 


find 


PLANE  TRIGONOMETRY. 

COS  41°  13'=  .7522233, 
cos  41°  14'=  .7520316; 
cos  41°  13'  26". 

cos  47°  38'=  .6738727, 
cos  47°  39'=  .6736577; 
cos  47°  38' 30". 


Ans.  .7521403. 


.6737652. 


74.  To  find  the  Angle  whose  Sine  is  Given. 

Find  the  angle  whose  sine  is  .5082784,  having  given  from 

sin  30°  33'  =  .5082901 

sin  30°  32' =  .5080396    '  '    . 


the  table 


diff.  for  1'  =  .0002505 

given  sine  =  .5082784 
sin  30°  32'  =  .5080396 


diff.  =  .0002388 

Let  d  =  diff.  between  30°  32'  and  required  angle ;   then 
.0002505  :  .0002388  :  :  60  :  d. 


d  = 


2388  X  60  _  6552 
2505 


167 
=  57.2  nearly. 

.♦.  required  angle  =  30°  32' 57".2. 

75.  To  find  the  Angle  whose  Cosine  is  Given. 

Kind  the  angle  whose  cosine  is  .40432§1,  having  given 


lioni  the  ^^,ble 


cos  66°  9' =  .4043436 
cos  66°  10'  =  .4040775 


',.'■-' 


diff.  for  1'  =  .0002661 

cos  66°  9' =  .4043436 
given  cosine  =  .4043281 

diff.  =  .0000155 


>  k 


EXAMPLES. 


109 


Let   d  =  diff.  between  66°  9'  and  required   angle ;   then 
.0002661 :  .0000155  :  :  60  :  d. 

2661 

Required  angle  is  greater  than  66°  9'  because  its  cosine  is 
/ess  than  cos  66°  9'.  ,'. 

.-.  required  angle  =  66°  9'  3".5.      .    - 


EXAMPLES. 

sin  44°  12' =  .6071651,  ' 

.      sin  44°  11'=  .6969565; 
whose  sine  is  .6970886.  Ans.  44'  11' 38". 

V       sin  48°  47' =  .7522233,  ' 

sin  48°  46' =  .7520316; 
whose  sine  is  .752140. 


1.  Given 

find  the  angle 

2.  Given 

find  the  angle 

3.  Given 


48°  46'  34" 


find  the  angle 

4.  Given 

find  the  angle 

5.  Given 

find  the  angle 

6.  Given 


sin  24°  11' =  .4096577, 
sin  24°  12'  =  .4099230 ; 
whose  sine  is  .4097559.    ; 

'"'      «os  32°  31' =.8432351, 
cos  32°  32' =  .8430787; 
whose  cosine  is  .8432. 

cos  44°  11'  =  .7171134, 
cos  44°  12' =  .7169106; 
whose  cosine  is  .7169848. 


24°  11'  22".2. 


32°31'13".5. 


44°  11'  38 


cos  70°  32'  =  .3332584, 
cos  70°  31' =  .3335326 ; 
find  the  angle  whose  cosine  is  .3333333. 


r- 


70°  31'  43".6 


110 


PLANE  TRIGONOMETRY. 


76.  Use  of  Tables  of  Logarithmic  Trigonometric  Func- 
tions. —  Since  the  sines,  cosines,  tangents,  etc.,  of  angles 
are  numbers,  we  may  use  the  logarithms  of  these  numbers 
in  numerical  calculations  in  \vhit;li  trigonometric  functions 
are  involved ;  and  these  logarithms  are  in  practice  much 
more  useful  than  the  numbers  themselves,  as  with  their 
assistance  we  are  able  to  abbreviate  greatly  our  calcula- 
tions ;  this  is  especially  the  case,  as  we  shall  see  hereafter, 
in  the  solution  of  triangles.  In  order  to  avoid  the  trouble 
of  referring  twice  to  tables — first  to  the  table  of  natural 
functions  for  the  value  of  the  function,  and  then  to  a  table 
of  logarithms  for  the  logarithm  of  that  function  —  the  log- 
arithms of  the  trigonometric  functions  have  been  calculated 
and  arranged  in  tables,  forming  tables  of  the  logarithms  of 
the  sines,  logarithms  of  the  cosines,  etc. ;  these  tables  are 
called  tables  of  logarithmic  sines,  logarithmic  cosines,  etc. 

Since  the  sines  and  cosines  of  all  angles  and  the  tangents 
of  angles  less  than  45°  are  less  than  unity,  the  logarithms  of 
these  functions  are  negative.  To  avoid  the  inconvenience  of 
using  negative  characteristics,  10  is  added  to  the  logarithms 
of  all  the  functions  before  they  are  entered  in  the  table. 
The  logarithms  so  increased  are  called  the  tabular  logarithms 
of  the  sine,  cosine,  etc.  Thus,  the  tabular  logarithmic  sine 
of  30°  is  ^  ^  ,      , 

10  +  log  sin  30°  =  10  -}-  log^  =  10  -  log  2  =  9.6989700. 

In  calculations  we  have  to  remember  and  allow  for  this 
in(;rease  of  the  true  logarithms.  When  the  value  of  any 
one  of  the  tabular  logarithms  is  given,  we  must  take  away 
10  from  it  to  obtain  the  true  value  of  the  logarithm. 

Thus  in  the  tables  we  find  ,  ■    . ' 

log  sin  31°  15' =  9.7149776. 

Therefore  the  true  value  of  tlie  logarithm  of  the  sine  of 
31°  15'  is  9.7149776  -  10  =  i.7l4977(;^ 

Similarly  with  the  logarithms  of  other  functions.    ^^'  * 


TABLES  OF  LOGARITHMIC  FUNCTIONS. 


Ill 


Note.  —  Eiigligh  authors  uauiilly  denote  tlieso  tabular  logarithms  by  the  letter  L. 
Thus,  IjsIii  a  denotes  the  tubular  loguriiliin  of  the  Bine  of  A. 

French  authors  use  the  logurithuis  of  the  tables  diminished  by  10.    Thus, 

log  sin  A  =  1.8598213,  instead  of  9.8598213. 

The  Tables  contain  the  tabular  logs  of  the  functions  of  all 
angles  in  the  first  quadrant  at  intervals  of  1' ;  and  from 
these  the  logarithmic  functions  of  all  other  angles  can  be 
found.* 

Since  every  angle  between  45°  and  90°  is  the  complement 
of  another  angle  between  45°  and  0°,  every  sine,  tangent, 
etc.,  of  an  cngle  less  than  45°  is  the  cosine,  cotangent,  etc., 
of  another  ingle  greater  than  45°  (Art.  16).  Hence  the 
degrees  at  th^^/  top  of  the  tables  are  generally  marked  from 
0°  to  45°,  and  those  at  the  bottom  from  45°  to  90°,  while 
the  minutes  are  marked  both  in  the  first  column  at  the  left, 
and  in  the  last  column  at  the  right.  Every  number  tliere- 
fore  in  each  column,  except  those  marked  diff.,  stands  for 
two  functions — the  one  named  at  the  top  of  the  column, 
and  the  complemental  function  named  at  the  bottom  of  the 
column.  In  looking  for  a  function  of  an  angle,  if  it  be  less 
than  45°,  the  degrees  are  found  at  the  top,  and  the  minutes 
at  the  left-hand  side.  If  greater  than  45°,  the  degrees  are 
found  at  the  foot,  and  the  minutes  at  the  right-hand  side. 

On  page  113  is  a  specimen  page  of  Mathematical  Tables. 
It  gives  the  tabular  logarithmic  functions  of  all  angles  between 
38°  and  39°,  and  also  of  those  between  51°  and  52°,  both 
inclusive,  at  intervals  of  1'.  The  names  of  the  functions 
for  38°  are  printed  at  the  top  of  the  page,  and  those  for  51° 
at  the  foot.  The  column  of  minutes  for  38°  is  on  the  left, 
that  for  51°  is  on  the  right. 

Thus  we  find 

log  sin  38°  29'  =9.7939907. 
log  cos  38°  45'  =  9.8920303. 
log  tan  51°  18'  =  10.0962856. 


', . 

jktSi^' 

'  i: 

Wl^f 

wm 

*  Many  tables  are  calculated  for  anglet  at  iotervaU  of  10". 


112 


PLANE  TRIGONOMETRY. 


^m 


.  :.;  ■  '■ 

*t 

^ 

iNl'  •  ■ 

i 

1      '  .;    iL.  ' 

r  iir  t-'-  ■■■' 

!l     M    :     ■  ■, 

I 

■        1 

:i      .1.: 

I  m 


i: 


77.  To  find  the  Logarithmic  Sine  of  a  Given  Angle. 

Find  log  sin  38°  52'  40". 
We  have  from  page  113 

log  sin  38"  53'  =  9.7977775 
log  sin  38°  52'  =  9.797()2()8 

diff.  for  1'=    .0001567  '  -.^' 

Let  d  =  diff.  for  46",  and  assuming  that  the  change  in 

the  log  sine  is  proportional  to  the  change  in  the  angle,  we 

have 

60  :  46  ::  .0001567  :  d. 

...  d  =  4^_x_M>lf!^=. 0001201.  r         •     .     . 

60 

.-.     log  sin  38°  52' 4.6"  =0.7976208 +  .0001201 

=  9.7977409. 

78.  To  find  the  Logarithmic  Cosine  of  a  Given  Angle. 

Find  log  cos  83°  27'  23",  having  given  from  the  table  ^ 

log  cos  83°  27'=  9.0571723  ;  • .  "-■:  :    ■■  ' 

log  cos  83°  28'  =  9.0560706 

diff.  fori' =    .0011017  '         ■ 

Let  d=  decrease  of  log  cosine  for  23";  then 

60  :  23  ::  .0011017  :  d.  .     .:  -:  ^  .    <- 

23  X  .0011017 


•.   d  = 


60 


=  .0004223,  nearly. 


.-.     log  cos  83°  27'  23"=  9.0571723  -  .0004223 

=  9.0567500. 

EXAMPLES. 

1.    Given      log  sin   6°  33'=  9.0571723,  { 


find 


log  sin    6°  32'=  9.0560706: 

log  e;«      fi°  ^'}'  'A7II 


sin    6°  32' 37 


Ans,  9.05675. 


\. 


38  Peg. 


TABLE  OF  LOGARITHMS. 


113 


Sine. 

9.7893420 
9.7895036 
9.7896652 
9.7898266 
9.7899880 
.?790I493_ 
9.7903104 
7  i  97904715 


8 

9 
10 


II 
12 


13 
IS 


23 
24 
25 
26 
27 
28 
29 
30 

31 
32 
33 
34 
35 
36 
37 
38 

39 
40 

41 

42 

43 
44 
45 
46 

47 
48 

49 
50 

51 

52 
53 
54 
55 
56 
57 
58 

59 
60 

f 


9.7906325 
9.7907933 
9.7909541 


9.791 1 148 
9.7912754 
9-7914359 
97915963 
9.7917566 

16  9.7919168 

17  I  9.7920769 

18  j  9.7922369 

19  I  97923968 

20  97925566 

21  '  9.7927163 

22  I  9.7928760 


97930355 
97931949 
97933543 


97935135 
97936727 
97938317 
97939907 
^.7941496 

9 

9 

9 

9 

_9 

9 
9 
9 
9 
9 


7943083 
,7944670 
,7946256 
,7947841 

79494?S 
,7951008 
7952590 
795417 I 
7955751 
7957330 

7958909 
,7960486 
,7962062 
,7963638 
7965212 

,7966786 

7968359 
,7969930 
,7971501 
_797397i 
9.7974640 
9.7976208 
9.7977775 

97979341 
9.7980906 


9.7982470 
9.7984034 

97985596 
9.7987158 
9.7988718 


Diff. 

616 

6x6 
614 
614 
613 
611 

611 
610 
608 
608 
607 

606 
60s 
604 
603 
602 

601 
600 

599 
598 
597 
597 
595 
594 
594 
592  i 

592 
590 
590 
589 
587 

587 
586 

585 
584 
583 
582 

581 
580 

579 
579 

577 
576 
576 
574 
574 

573 
571 
571 
570 
569 
568 

567 
566 

565 
564 

564 
562 
562 
560 


Tang.   I  Diff.  {  Ootang. 


Diff. 


9.8928098 
9.8930702 
9-8933306 
9.8935909 
9.893851 1 
9.8941 114 

9-8943715 
9.8946317 
9.8948918 
9.8951519 
9.89541 19 

9.8956719 

9-8959319 
9.8961918 
9.8964517 
9.89671 16 


9.8969714 
9.8972312 
9.8974910 
9.8977507 
9.8980104 


9.8982700 
9.8985296 
9.8987892 
9.8990487 
9.8993082 

9-8995677 
9.8998271 
9.9000865 
9.9003459 
9.9006052 

9.900864s  i 
9.901 1237  I 

9.9013830  ; 
9.9016422 

9-90}9pj3  ' 
9.9021604  I 
9.902419s 

9.9026786  ; 
9.9029376  /I 
9.9031966 

9-9034555  ' 
9.9037144 

9-9039733 
9.9042321 
9.9044910 


9.9047497 
9.9050085 
9.9052672 

9-9055259 
9.9057845 


9.9060431 
9.9063017 
9.9065603 
9.9068183 
9^9070773 

9-9073357 
9.9075941 
9.9078525 
9.908 1 109 
9.9083692 


Oosine.     |  Diff.  !    Ootang. 


2604 
2604 
2603 
2602 
2603 
2601 

2602 
2601 
2601 
2600 
2600 

2600 
2599 
2599 
2599 
2598 

2598 
2598 

2597 
2597 
2596 

2596 
2596 
2595 
2595 
2595 

2594 
2594 
2594 
2593 
2593 
2592 

2593 
2592 
2591 
2591 

2591 
2591 
2590 

2590 
2589 

2589 
2589 
2588 
2589 
2587 
2588 
2587 
2587 
2586 
2586 

2586 
2586 
2585 
2585 
2584 

2584 
2584 
2584 
2583 


0,1071902 
0.1069298 
0.1066694 
0.1 06409 1 
0.1061489 
0.1058886 

0.1056285 
0.1053683 
0.1051082 
0.104848 1 
0.1045881 


0.1043281 
0.104068 1 
0.1038082 
0.1035483 
0.1032884 
0.1030286 
0.1027688 
0.1025090 
0.1022493 
0.1019896 


0.1017300 
0.1014704 
0.1012108 
0.1009513 
0.1006918 


0.1004323 
0.1001729 
0.099913s 
0.0996541 
0-0993948 


0-0991355 
0.0988763 
0.0986170 
0.0983578 
0.0980987 


0.0978396 
0.0975805 
0.0973214 
0.0970624 
0.0968034 


0.0965445 
0.0962856 
0.0960267 
0.0957679 
0.0955090 


0.0952503 
0.0949915 
0.0947328 
0.0944741 
0.0942155 


0.0939569 
0.0936983 
0.0934397 
0.0931812 
0.0929227 


0.0926643 
0.0924059 
0.0921475 
0.0918891 
0.0916308 


Diff.  i   Tang. 


987 
988 
988 
989 
990 
990 

991 
992 
992 
992 
993 

994 
995 
995 
995 
997 

996 
998 
998 
998 
1000 

999 
001 
001 
001 
003 

002 
004 
004 
004 

005 

006 
007 
007 
007 
008 

009 
010 
010 
010 
on 

012 
013 
013 
013 
014 

015 
016 
016 
oi6 
018 

017 
019 
019 
020 

020 
021 
021 
022 
023 


Oosine. 


9.8965321 
9.8964334 
9.8963346 
9.8962358 
9.8961369 
9.8960379 


9.8959389 

9.8958398 
9.8957406 
9.8956414 
_9:8955422 
9.8954429 

9-8953435 
9.8952440 
9.8951445 
9.8950450 

9.8949453 
9-8948457 
9.8947459 
9.8946461 
9.8945463 


9.8944463 
9.8943464 
9.8942463 
9.8941462 
9.8940461 


9-8939458 
9.8938456 
9.8937452 
9.8936448 
9.8935444 


9.8934439 

9-8933433 
9.8932426 
9.8931419 
9.8930412 


9.8929404 
9-8928395 

9-8927385 
9.892637s 
9.8925365 


9-8924354 
9.8923342 
9.8922329 
9.8921316 
9.8920303 


9.8919289 
9.8918274 
9.8917258 
9.8916242 
9.8915226 


Diff. 


9.8914208 
9.8913191 
9.8912172 
9.8911153 
9.8910133 

9.8909113 
9.8908092 
9.8907071 
9,8906049 
9.8905026 


60 
59 
58 
57 
56 
55 
54 
53 
52 
51 
50 

49 
48 

47 
46 
45 
44 
43 
42 

I  41 
J40 

39 
38 
37 
36 
35 
34 
33 
32 
31 
30 

29 
28 
27 
26 

25 

24 

23 
22 
21 
20 

19 
18 

17 
16 

IS 

14 

I  13 

!   12 

i  11 

10 

9 
8 

7 
6 

5 

4 
3 
2 
I 
o 


Sine. 


f 


■  tf'-: 


: 

51  Degr. 


114 


If 


2. 

Given 

find 

3. 

Given 

find 

4. 

Given 

find 


PLANE  TRIGONOMETRY. 

log  sin  55"  33'=  i).91()2530, 
log  sin  55°  34'=  U.U1G34()G ; 
log  sin  55°  33'  54". 

log  cos  37°  28'=  9.8()0(>604, 
log  cos  37°  29'=  9.8995636 ; 
log  cos  37°  28' 36". 

log  cos  44°  35'  20'=  9.8525789, 
log  cos  44°  35'  30"=  9.8525582 ; 
logcos44°35'25".7. 


Ans.  9.9163319. 


9.8996023. 


9.8525671. 


See  foot-note  of  Art.  76. 

5.  Given  log  cos  55°  1 1 '  =  9. 7565999, 
log  cos  55°  12'=  9.7561182; 
logcos  55°  11' 12". 

logtan27°  13'=  9.7112148, 
log  tan  27°  14'=  9.7115254  ; 
log  tan  27°  13' 45". 


find 

6.   Given 

find 


9.7565636. 


9.7114477. 


79.   To  find  the  Angle  whose  Logarithmic  Sine  is  Oiven. 

Find  the  angle  whose  log  sine  is  8.8785940,  having  given 

from  the  table 

logsin4°  21'=  8.8799493 
log  sin  4°  20'=  8.8782854 

diff.  forl'=    .0016639 

given  log  sine  =  8.8785940 
log  sin  4°  20'=  8.8782854 

diff.  =  ~T0003086 

Let  d  =  diff.  between  4°  20'  and  required  angle  ;  then 

.0016639  :  .0003086  :  :  60  :  d. 

,      3086  X  60      .- ,  , 

.  •,    d= ~ —  =  24,  nearly. 

16639  '  ^ 

.'.  required  angle  =  4°  20'  24". 


EXAifPLES. 


116 


80.  To  find  the  Angle  whose  Logarithmic  Cosine  is  Given. 

Find  the  angle  whoso  log  cosine  is  9.8934342. 
We  have  from  page  113 

log  cos  38°  31'=  9. 8934430 
log  cos  38°  32'  =  9.8933433 

cliff,  for  1'=    .0001006 

log  cos  38°  31'=  9.8934439 
given  log  cosine  =  9.8934.342 

cliff.  =    .0000097 
Let  d  =  (liff.  between  38°  31'  and  required  angle ;  then 

.OOOIOOG  :  .0000097  ::  GO  :  d.     "      "    '"' 

.^^^)00O()97       .,^97x«0^ 
.0001006  1006 

.-.  required  angle  =  38°  31'  5".8. 

Note.  — In  using  both  the  tables  of  the  nntural  sines,  cosines,  etc.,  and  the  tables 
of  tlie  loffarithmk;  sines,  cosines,  etc.,  the  stiuient  will  remember  that,  in  the  first 
quadrant,  as  the  uiikIu  increases,  the  sine,  taugvMit,  and  secant  increase,  but  the 
cosine,  cotangent,  and  cosecant  decrease. 

EXAMPLES.        ,      V     ':::,■..■•.:■'■' 

1.  Given        log  sin  14°  24'=  9.3956581, 

log  sin  14°  25'=  9.3961499 ; 
find  the  angle  whose  log  sine  is  9.3959449.    Ans.  14°  24' 35". 

2.  Given        log  sin  71°  40'=  9.9773772, 

log  .sin  71°  41'=  9.9774191; 
find  the  angle  whose  log  sine  is  9.9773897.  71°  40'  18". 

3.  Given        log  cos  28°  17'=  9.9447862, 

log  cos  28°  16'=  9.9448541 ; 
find  the  angle  whose  log  cosine  is  9.9448230.      28°  16'  27".5. 


It 


~»t 


V 


116 


PLANE  TRIGONOMETRY. 


4.   Given        log  cos  80°  Tni'  =  '.).  1<K)871)3, 

log  cos  80°  52'  50"  =  9.2000105 ; 
find  the  angle  whose  log  cosine  is  9.2000000. 

Ans.  80°  52' 51". 


5,  Given        log  tan  35°  4'=  9.8463018, 

log  tan  35°  5'=  9.8465705; 
find  the  angle  whose  log  tangent  is  9.8464028. 

6.  Given        log  sin  44°  35' 30"=  9.8463678, 

log  sin  44°  35'  20"=  9.8463464; 
iind  the  angle  whose  log  sine  is  9.8463586. 


35°  4'  23' 


44°  35'  25".7. 


7.    Find   the   angle   by   page   113  whose   log  tangent   is 
10.1018542.  Ans.  51°39'28".7. 


81.  Angles  near  the  Limits  of  the  Quadrant.  —  It  was 

assumed  in  Arts.  72-80  that,  in  general,  the  differences  of 
the  trigonometric  functions,  both  natural  and  logarithmic, 
are  approximately  proportional  to  the  differences  of  their 
corresponding  angles,  with  certain  exceptions.  The  excep- 
tional cases  are  as  follows : 

(1)  Natural  functions. — For  the  sine  the  differences  are 
insensible  for  angles  near  90°;  for  the  cosine  they  are  in- 
sensible for  angles  near  0°.  For  the  tangent  the  differences 
are  irregular  for  angles  near  90°;  for  the  cotangent  they  are 
irregular  for  angles  near  0°. 

(2)  Logarithmic  functions. — The  principle  of  propor- 
tional ])arts  fails  both  for  angles  near  0°  and  angles  near 
90°.  For  the  log  sine  and  the  log  cosecant  the  differences  are 
irregular  for  angles  near  0°,  and  insensible  for  angles  near  90°. 
For  the  log  cosine  and  the  log  secant  the  differences  are  in- 
sensible for  angles  near  0°,  and  irregular  for  angles  near  90°. 
For  the  log  tangent  and  the  log  cotangent  the  differences  are 
irregular  for  angles  near  0°  and  angles  near  90°. 


EXAMPLES. 


117 


It  follows,  therefore,  that  angles  near  0°  and  angles  near 
90°  cannot  be  found  with  exactness  from  their  log  trigono- 
metric functions.  These  difficulties  may  be  met  in  three 
ways. 

(1)  For  an  angle  near  0°  use  the  principle  that  the  ainea 
and  tangents  of  small  angles  are  approximately  proiiortional 
to  the  angles  themselves.     (See  Art.  130.) 

(2)  For  an  angle  near  90°  use  the  half  angle  (Art.  99). 

(3)  In  using  the  proportional  parts,  find  two,  three,  or 
more  orders  of  differences  (Alg.,  Art.  197). 

Special  tables  are  employed  for  angles  near  the  limits  of 
the  quadrant.  > 

...  BXAMPLBS, 

1.  Given  log,,,  7  =  .8450980,  find  log,,, 343,  log,„ 2401,  and 
logio  16.807.  Ans.  2.5352940,  3.3803920,  1.2254900. 

2.  Find  the  logarithms  to  the  base  3  of  9,  81,  \,  ^j,  .1,  -^j. 

Ans.  2,  4,  -  1,  -  3,  -  2,  -  4. 

3.  Find  the  value  of  log2  8,  loga.5,  logy  243,  log5(.04), 
logiolOOO,  log,o  .001.  Ans.  3,  -  1,  5,  -  2,  3,  -  3. 

4.  Find  the  value  of  log„  a*,  \og(,Vb%  logs 2,  log^;  3,  logmolO. 

^ns.  I,  I,  I,  I,  |. 

Given  log,o  2  =  .3010300,  log,o  3  =  .4771213,  and  logio7== 

.8450980,  find  the  values  of  the  following : 

.  . '    .,..,-...,*•=  *•••.■,»• 

.  5.   Iogio35,  logiul50,  logio.2. 

Ans.  1.544068,  2.1760913,  1.30103. 

6.  logio  3.5,  log,„  7.29,  log,o. 081.  _ 

Ans.  .5440680,  .8627278,  2.9084852. 

7.  log,o|,  log,„3^1og,o\/-i?. 

Ans.  .3679767,  2.3856065,  .0780278. 


J. 


118 


PLANE  TiilGONOMETRY. 


8.  Write  down  the  integral  })ai't  of  the  common  loga- 
rithms of  796o,  .1,  2.61,  79.6341,  1.0006,  .00000079. 

Ans.  3,-1,  0,  1,  0,  -  7. 

9.  Give  the  position  of  the  first  significant  figure  in  the 
numbers  whose  logarithms  are 

2.4612310,  1.2793400,  6.1763241. 

10.  Give  the  position  of  the  first  significant  figure  in 
the  numbers  whose  logarithms  are  4.2990713,  .3040595, 
2.5860244,  3.1760913,  1.3180633,  .4980347. 

Ans.  ten  thousands,  units,  hundreds,  3rd  dec.  pi.,  1st 
dec.  pi.,  units.  ,   .  . 

11.  Given  log  7  =  .8450980,  find  the  number  of  digits  in 
the  integral  part  of  7'",  49«,  343'"",  (Y)^",  (4.9)'^  (3.43)'". 

Ans.  9,  11,  85,  4,  9,  6. 

12.  Find  the  position  of  the  first  significant  figure  in  the 
numerical  value  of  20^,  (.02)^,  (.007)^  (3.43) ''o,  (.0343)«, 
(.0343)t'5.      .  , 

Ans.  tenth  integral  pi.,  12th  dec.  pi.,  5th  dec.  pi.,  units, 
12th  dec.  pi.,  1st  dec.  pi. 

Show  how  to  transform 

13.  Common  logarithms  to  logarithms  with  base  2. 

Ans.  Divide  each  logarithm  by  .3010."). 

14.  Logarithms  with  base  3  to  common  logarithms. 

Ans.  Multiply  each  log  by  .4771213. 

15.  Given  logio2  =  ,3010300,  find  logglO.  3.32190. 

16.  Givenlog,o7  =  .S450980,  findlog^lO.        \  1.183. 

17.  Given  log,o  2=-  .3010300,  find  logs  10.  1.10730. 

18.  The  mantissa  of  the  log  of  85762  is  9332949;  find 
(1)  the  log  of  a/.0085762,  and  (2)  the  number  of  figures  in 
(85762)",  when  it  vi  multiplied  out. 

Ans.  (1)  1.8121177,  (2)  55. 


EXAMPLES. 


119 


19.  What   are   the   characteristics  of  the  logarithms  of 
3742  to  the  bases  3,  G,  10,  and  12  respectively  ? 

.     ^;l     ■  /     :  Ans.  7,  4,  3,  3. 

20.  Prove  that  7  log  f|  +  G  log  |  +  5  log  f  +  log  || = log  3. 

1 


Aus.  2  + 


21.  Given  log.o  7,  find  log,  490. 

22.  From  5.3429  take  3.G284. 

23.  Divide  13.2G15  by  8. 

24.  Prove  that  G  log  |  +  4  log  ^\  +  2  log  -\^-  =  0 

25.  Find  log  |297VIl| » to  the  base  3Vn. 

Given  log  2  =  .3010300,  log  3  =  .4771213. 


logio  7 
3.7145. 

2.407G. 


1.8. 


2G.    Fijid  log21G,  G480,  5400,  A 

Ans.  2.3344539,  3.8115752,  3.7323939,  r.G478174. 

27.  Find  log  .03,  G"',  (5^)"^. 

Ans.   2.4771213,  I.740G1G2,  1.G36500G. 

28.  Find  log. 18,  log 2.4,  log^^g-. 

,    A71S.  i.255272G,  .3802113,  1.2730013. 

29.  Find  log  (G.25)^,  log4VJ005.  .113G971,  1.45154. 


log  5G321  =  4.750G704, 
log  56322  =  4.750G781; 
log  5632147.  6.7506740. 

log  53403  =  4.7275657, 

log  53402  =  4.7275575; 

log  5340234.  6.7275603. 

log564l2  =  4.7513715, 

log  56413  =  4.7513792 ; 

log  564.123.  2.7513738. 


30. 

Given 

find' 

31. 

Given 

find 
32. 

Given 

find 


120 


PLANE  TRIGONOMETRY. 


33. 

Given 

find 

34. 

Given 

find 

35. 

Given 

find 

36. 

Given 

find 

37. 

Given 

find 

:■  •    ■■    ., 

38. 

Given 

find 

39. 

Given 

iind  the  number 

40. 

Given 

find  the  number 

41. 

Given 

find  the  number 

Ans.   4.9413333. 


log  87364  =  4.9413325, 
log  87365  =  4.9413375; 
log  .0008736416. 

log  37245  =  4.5710680, 

log  37246  =  4.5710796; 

log  3.72456.  .5710750. 

log  32025  =  4.5054891, 

log  32026  =  4.5055027; 

log  32.025613.  1.5054974. 

log  65931  =  4.8190897, 

log  65932  =  4.8190962;  ■  ; 

log  .000006593171.  _^.,  6.8190943. 

log  25819  =  4.4119394,   ;  ' 

log  25820  =  4.4119562 ;    :   ;  x  %> 

log  2.581926.  .4119438. 

log 23454  =  4.3702169,   ,  ft  ;    *:: 

log  23453  =  4.3701984; 

log  23453487.  ■  ,    v  ,  ^  T.3702074. 

log  45740  =  4.6602962,     ^  .  : : '  .v> 
log  45741  =  4.6603057; 
whose  logarithm  is  4.6602987.   45740.26. 

log  43965  =  4.6431071,     -      r 
log  43966  =  4.6431170;         ^  ' 
whose  logarithm  is  4.6431150.  .000439658. 

log  ^6891  =  4.7550436, 
log  56892  =  4.7550512; 
whose  logarithm  is  .7550480.    5.689158. 


m 


EXAMPLES. 


121 


42.  Given  log  34572  =  4.5387245, 

log  34573  =  4.5387371; 
find  the  number  whose  logarithm  is  2.5387359. 

Alls.  345.7291. 

43.  Given  log  10905  =  4.037G257,  , 

log  10906  =  4.0376655 ; 
find  the  number  whose  logarithm  is  3.0376371.      1090.5286. 

44.  Given  log  25725  =  4.4103554, 

log  25726  =  4.4103723; 
find  the  number  whose  logarithm  is  7.4103720. 

Alls.  .00000025725982. 

In  the  following  six  examples  the  student  must  take  his 
logarithms  from  the  tables. 

45.  Required  the  prciuct  of  3670.257  and  12.61158,  by 
logarithms.  '        Aiis.  46287.74. 

46.  Required  the  quotient  of  .1234567  by  54.87645,  by 
logarithms.  Ans.  .002249721. 

47.  Required  the  cube  of  .3180230,  by  logarithms. 

Ans.  .03216458. 

48.  Required  the  cube  root  of  .3663265,  by  logarithms. 

A71S.  .7155316. 


49.  Required  the  eleventh  root  of  63.742. 

50.  Required  the  fifth  root  of  .07. 

51.  Given  sin  42°  21'=  .6736577, 

•    sin  42°  22' =.6738727; 
■  '■-  sin  42°  21' 30". 


1.45894. 
.68752. 


find   -''■      • 
52.   Given 

find 


.6737652. 


sin  67°  22'=  .9229865, 
sin  67°  ^3'  =  .9230984 ; 
sin  67°  22'  48".6. 


.9230769. 


122 


PLANE  TRIGONOMETRY. 


53. 

Given 

find 

54. 

Given 

find 
55. 

Given 

find 


sin   7°  17' =  .1267761, 
sin   7°  18' =  .1270646; 
sin   7"  17' 25". 

cos  21°  27'  =  .9307370, 
cos  21°  28'  =  .9306306 ; 

cos  21°  27' 45". 

cos  34°  12'  =  .8270806, 
cos  34°  13' =  .8269170; 

cos  34°  12'  19".6. 


Ana.  .1268963. 


.9306572. 


.8270272. 


66.    Given 

sin  41°  48' =  .0665325, 

■-■■•' 

sin  41°  49' =  .6667493; 

.                  find  the  angle  whose  sine  is  .6666666. 

67.   Given 

sin  73°  44'  =  .9599684, 

sin  73°  45' =  .9600499; 

find  the  angle 

whose  sine  is  .96. 

58.   Given 

cos  75°  32' =  .2498167, 

■'     ^     1,  . 

cos  75°  31' =.2500984; 

find  the  angle 

whose  cosine  is  .25. 

!                      69.    Given 

cos  53°    7' =  .0001876, 

cos  53°    8' =  .5999549; 

find  the  angle 

whose  cosine  is  .6. 

60.   Given 

log  sin  45°  16'  =  9.8514969, 

1 

log  sin  45°  17'  =  9.8516220 ; 

'                  find 

log  sin  45°  16' 30". 

61.    Given 

log  sin  38°  24'  =  9.7931949, 

log  sin  38°  25'  =  9.7933543  ; 

find 

log  sin  38°  24'  27". 

41°  48'  37 


73°  44'  23".2. 


76°  31' 21' 


63°  T  48".4. 


9.8515594. 


9.7932666. 


EXAMPLES. 


123 


62. 

Given 

find 

03. 

Given 

find 

64. 

Given 

find 

65. 

Given 

find 

66. 

Given 

find 

67. 

Given 

find 

68. 

Given 

find 

69. 

Given 

find 

■■  . 

70. 

Given 

log  sin  32°  28' =  9.7298197, 
log  sin  32°  29'  =  9.7300182 ; 
log  sin  32°  28' 36".  Ans.  9.7299388. 

log  sin  17°  1'  =  9.4603483. 
log  sin  17°  0'  =  9.4659353  J 
log  sin  17°    0'12".       ' 


find 


log  sin  26°  24'  =  9.6480038. 
log  sin  26°  25'  =  9.6482582; 
log  sin  26°  24' 12". 

log  cos  17°  31'  =  9.9793796, 
log  cos  17°  32'  =  9.9793398 ; 
logcosl7°31'25".2. 

log  tan  21°  17'  =  9.5905617, 
log  tan  21°  18'  =  9.5909351 ; 
log  tan  21°  17'  12". 

log  tan  27°  26'  =  9.7152419, 
log  tan  27°  27' =  9.7155508; 
log  tan  27°  26'  42". 

log  cot  72°  15'  =  9.5052891, 
log  cot  72°  10'  =  9.5048538 ; 
log  cot  72°  15' 35". 

log  cot  36°  18'  =  1 0.1339650, 
log  cot  36°  19'  =  10.1337003 ; 
log  cot  36°  18'  20". 

log  cot  51°  17'  =  9.9039733, 
log  cot  51°  18'  =  9.9037144 ; 
log  cot  51°  17' 32". 


9.4660179. 


9.6480547. 


9.9793629. 


9.5906364. 


9.7154581. 


9.5050352. 


10.1338768. 


9.9038352. 


124 


PLANE  TRIGONOMETRY. 


■'. ' 

■    .r.i:i 

il 

I  tvyr,-,- 

illf 

i     i 

71.   Given       log  sin  16°  19' =  9.4486227, 
log  sin  16°  20'  =  9.4490540 ; 
find  the  angle  whose  log  sine  is  9.4488105. 


Ans.  16°  19' 26". 


72.  Given      log  sin   6°  53'       =9.0786310, 

log  sin   6°  53'  10"=  9.0788054 ; 
find  the  angle  whose  log  sine  is  9,0787743. 

73.  Given      logcos  22°  28' 20"=  9.9657025, 

log  cos  22°  28'  10"=  9.9657112 ; 
find  the  angle  whose  log  cosine  is  9.9657056. 


6°  53'  8 


'  CM 


22°  28' 16' 


In  the  following  examples  the  tables  are  to  be  used : 

74.  Find  log  tp.n  55°  37' 53".  Ans.  10.1650011. 

7^.  Findiogsin73°20'15".7.  9.9813707. 

76.  Find  log  cos  55°  11' 12".  9.7665636. 

77.  Find  log  tan  16°    0'2"r'.  9.4577109. 

78.  Find  log  sec  16°    0'27".  10.0171747. 

79.  Find  the  angle  whose  log  cosine  is  9.9713383. 

Ans.  20°  35' 16". 

80.  Find  the  angle  whose  log  cosine  is  9.9165646. 

Ans.  34°  23' 25". 


81.  Find  log  cos  34°  24' 36". 

82.  Find  log  cos  37°  19'  47". 

83.  Find  log  sin  37°  19'  47". 

84.  Findlogtan37°19'47". 

85.  Find  log  sin     >°  18' 24".6. 

86.  Find  log  cos  32°  18'  24".6. 

87.  Find  log  tan32°  18'  24".6. 


9.9164762. 
9.9004540. 
9.7827599. 
9.8823059. 
9.7279096. 
9.9269585. 
9.8009511. 


EXAMPLES. 


125 


Prove  the  following  by  the  use  of  logarithms ; 

88     (7014)'^- 1  ^  .9942207. 
(7.014)«+1 


gg    j^5:g_xj^00307j  ^  .000232432. 


90.    W(2002)^'^'»x(i001g_9is^o.'^00000. 
\         1001  X  2002  • 


Vfc 


126 


PLANE   TRIGONOMETRY. 


CHAPTER  V. 

SOLUTION  or  TRiaONOMETEIO   EQUATIONS. 

82.  A  Trigonometric  Equation  is  an  equation  in  which 
the  unknown  quantities  involve  trigonometric  functions. 

The  solution  of  a  trigonometric  equation  is  the  process  of 
finding  the  vahies  of  the  unknown  quantity  which  satisfy  the 
equation.  As  in  Algebra,  we  may  have  two  or  more  simul- 
taneous equations,  the  number  of  angles  involved  being 
equal  to  the  number  of  equations. 


EXAMPLES. 


1.    Solve  sin^  = 


^ 


This  is  a  trigonometric  equation.     To  solve  it  we  must 


find  some  angle  whose  sine  is 


We  know  that  sin  30°=  1- 

2 

Therefore,  if  30°  be  put  for  6,  the  equation  is  satisfied. 
.-.  $z=  30°  is  a  solution  of  the  equation. 

.-.  ^  =  n7r4-(-l)"- (Art.  38) 

6 

5 

2.   Solve  cos  ^  4-  sec  ^  =  -• 

The  usual  method  of  solution  is  to  express  all  the  func- 
^■ions  in  terms  of  one  of  them. 

Thus,  we  put  -         for  sec^,  and  get 
cos  9 


1 


cos^  +  — —  =  - 
cos^     2 


TRIGONOMETRIC  EQUATIONS. 


127 


This  is  an  equation  in  which  9,  and  therefore  cos^,  is 
unknown.  We  proceed  to  solve  the  equation  algebraically 
just  as  we  should  if  x  occu])ied  the  place  of  cos^,  thus : 


CQS'O 


^  cos  6  =  —  1. 

'.  cos^  =  -±^ 
4     4 


2  or  -. 


1 

2 


The  value  2  is  inadmissible,  for  there  is  no  angle  whose 
cosine  is  numerically  greater  than  1  (Art.  21). 


cos  ^  =    • 


But 


cos  60°  = 


2' 


:;  ;  -  .-.  cos^  =  cos60°. 

Therefore  one  value  of  6  which  satisfies  the  equation  is  60°. 

3.  Solve    cosec  6  -  cot'^^ -f- 1  =  0, 

We  have  cosec  0  -  (cosec^  ^  -  1 )  +  1  =  0 .  .  (Art.  23) 
cosec^^  —  cosec  ^  =  2. 

.'.  cosec^  =  -  ±  - 

Z  Li 

=  2  or  -1.  .     ; 

But  cosec  30°  =  2. 

.'.  cosec  ^  =  cosec  30°. 
Therefore  30°  is  one  value  of  0  which  satisfies  the  equation . 

Find  a  value  of  B  which  will  satisfy  the  following  equa- 
tions: 

4.  cos^  =  cos2<?.  Ans.  t-n. 

5.  2 cos 6  =  seed.  /    46°. 


128 


PLANE  TRIGONOMETRY. 


6. 

4sin^  — 3cosec^  =  0. 

Ans.  60": 

7. 

4  cos  6  ='6  sec  6. 

30°. 

8. 

3. sill  ^-2008=^^  =  0. 

30°. 

9. 

V2  sin  6  =  tan  0. 

0°  or  45°. 

10. 

tan  <?  =  3  cot  0. 

60°. 

11. 

tan  ^  +  3  cot  ^  =  4. 

45°. 

12. 

cos  0  +  cos  3  ^  +  cos  o 

9  =  0. 

o 

7r    Jtt 

2'   3 

13. 

sin  [0  —  (f>)  =  ^,  cos  (6 

4-c^)  =  0. 

^  =  60 

",</»  =  30°. 

83.   Solve  the  equations 

7)1  am  <l>  =  a (1) 

m  cos  </)  =  6 (2) 

where  a  and  6  are  given,  and  the  values  of  m  and  <f>  are 
required. 

Dividing  (1)  by  (2),  we  get  *       '  -^^ ' 

tan  </>=-, 
ft 

which  gives  two  values  of  </>,  differing  by  180°,  and  there- 
fore two  values  of  m  also  from  either  of  the  equations 

b 


m  = 


a 


sin  cf>      cos  <^ 

The  two  values  of  m  will  be  equal  numerically  with 
opposite  signs. 

In  practice,  m  is  almost  always  ^jositive  by  the  conditions 
of  the  problem.  Accordingly,  sin  <^  has  the  sign  of  a,  and 
cos  <!>  the  sign  of  6,  and  hence  <^  must  be  taken  in  the  quad- 
rant denoted  by  these  signs.  These  cases  may  be  considered 
as  follows : 

(1)  Sin  (f>  and  cos  </>  both  positive.  This  requires  that  the 
angle  <^  be  taken  in  the  first  quadrant,  because  sin  «^  and 
cos  ^  are  both  positive  in  no  other  quadrant. 


or 


or 


TRIGONOMETRIC  EQUATIONS. 


129 


(2)  Sin  ff)  positive  and  cos  <^  negative.  This  requires  that  <fi 
be  taken  in  the  second  quadrant,  because  only  in  this  quad- 
rant is  sin  <^  positive  and  cos  <^  negative  for  the  same  angle. 

(3)  Sin  <f)  and  cos  <^  both  negative.  This  requires  that  <fi 
be  taken  in  the  third  quadrant,  because  only  in  this  quad- 
rant are  sin  <^  and  cos  ^  both  negative  for  the  same  angle. 

(4)  Sin  <^  negative  and  cos  <f>  j)(>sitive.  This  requires  that 
<f>  be  taken  in  the  fourth  quadrant,  because  only  in  this 
quadrant  is  sin<^  negative  and  cos  t^  positive  for  the  same 
angle. 

Ex.  1.  Solve  the  equations  msin</)  =  332.76,  and  mcos  <^ 
=  21)0.08,  for  m  and  <\>. 

log  m  sin  <^  =  2.52213 

-     r     ■■     '^  log  m  cos  <^  =  2.46252  .      .    " 


<^  =  48°  55'.2. 


m  =  441.45. 


log  tan  <^  =  0.05061 

log  m  sin  <^  =  2.52213 
log  sin  <i>  =  0.87725 

log  m  =  2.64488 


Ex.  2.    Solve  m  sin  <^  =  -  72.631,  and  m  cos  ^  =  38.412. 

Ans.  <f>  =  117°  52'.3,  m  =  -  82.164. 

84.   Solve  the  equation  ,  . 

a  sin  </>  +  6  cos  </>  =  c (1) 

a,  b,  and  c  being  given,  and  ^  required. 

Eind  in  the  tables  the  angle  whose  tangent  is  - ;  let  it 
be;8.  ^.  -:    ^  ^ 

Then  -=tan/8,  and  (1)  becomes 

a  ,  . 


or 


or 


"I 


a  (sin  <f)  -\-  tan  /?  cos  <^)  =  c ; 
/sin  <^  cos  13  +  cos  (f>  sin  (B\  _    , 


cos)3 


=  c; 


sin  (<i  -f  fl)  =  -  cos  ^  =  ^  sin  jS 
a  b 


(2) 


«■;  y, 


:,e 


•r^. 


■■!'* 


11 


ht 


130  PLANE  TRIGONOMETRY, 

There  will  be  two  solutions  from  the  two  values  of  ^  +  /3 
given  in  (2). 

Find  from  the  tables  the  value  of  cos  p.     Next  find  from 

the   tables   the  magnitude  of  the  angle  a  whose  sine  =  - 

a 
cos  p,  and  we  get 

sin  (</>  +  ;8)  =  sin  a, 

.-.  <^  +  /3  =  mtt +  (-!)'•«    .     .    (Art.  38) 
.-.  <^  =  _^-|-rt7r +  (-!)"«, 

where  n  is  zero  or  any  positive  or  negative  integer. 

In  order  that  the  solution  may  be  possible,  it  is  necessary 

to  have  -  cos  /8  =,  or  <  1. 
a 

NoTS. — This  example  might  bBve  been  solved  by  squaring  both  sides  of  the 
equation;  but  in  solving  trigonometric  equations,  it  is  important,  if  possible,  to 
avoia  squaring  both  sides  of  the  equation. 

Thus,  solve  cos 0  =  ka\nO (3) 

If  we  square  both  sides  wc  get 

cos2  0  =  A' slu2  fl  =  A»(l  -  cos*  fl). 

.'.  co8'9  =  -^^;  or  cosO=  ± — ^ —     ..........     (4) 

k 

Now  if  a  be  the  least  angle  whose  cosine  =  —  ,  we  get  from  (4) 

^'l  +  k^ 

«  =  nir  ±  a (5) 

But  (3)  may  be  written  cot  B  —  k. 

.'.  0  =  nn  +  a (6) 

(6)  is  the  complete  solution  of  the  given  equation  (3),  while  (5)  is  the  solution  of 
both  cos  fl  =  A;  sin  6,  and  also  of  cos  B  =  —  k  sin  9.  Therefore  by  squaring  both  nu-m- 
bers  of  an  equation  we  obtain  solutions  which  do  not  belong  to  the  given  equation. 

EXAMPLES. 

1.  Solve  0.7466898  sin  <^  -  1.0498  cos  <^  =  -  0.431689, 
when  <^  <  180°. 

log  6  =  0.02112-* 

log  a  =  1.87314 
log  tan /?  =  014798 - 
.-.  /3  =  125°  25' 20". 

*  The  minus  sign  is  written  thus  to  denote  that  it  belongs  to  the  natural  number 
and  does  not  affect  the  logarithm.  Sometimes  the  letter  n  is  written  instead  of  the 
minus  sign,  to  denote  the  same  thing. 


TBIGONOM-ETHIC  EQ  UA  TIONS. 


131 


log  sill  13  =  9.91111  ' 

log  c=  1.63517 - 
colog  6  =  9.97888- 

logsin(</)  +  y3)  =  9.52516+ 

,-.  <I>  +  I3=  19°  34'  40"  or  160°  25'  20". 
.-.  «^  =  -105°50'40"  or  35°0'0". 

2.  Solve  -  23.8  sin  <f>  +  19-3  cos  0  =  17.5 ((^  <  180"). 

Ans.  <^  =  4°  12'.7  or  -  106°  7'.9. 

3.  Solve  2  sin  e  +  2  cos  ^  =  V2.  Ans.  -  --\-nn  +  (-1)"-- 

4  6 


4.  "      sin0  +  V3cosd  =  l. 

5.  "      sin^  — cos0  =  1. 

6.  "      V3sin0-cos^  =  V2.         ^ir-{-7iir +  (-iy\ir. 


85.   Solve  the  equation 

sin  («  4- ic)  =  m  sino; 
in  which  a  and  m  are  given. 

From  (1)  we  have 

sill  («  4-  a?)  4-  si^l x  _m-\-l 


(1) 


sin  («  4-  a;)  —  sin  a;      m  —  1 

,    ■'  tanra;4-|j 

tan- 
ks',;' 2    -^ 

,      /     ,i\     wi4-li.     « 
.'.  tan  (a;  4- 4a)  =  — ^2__^tan- 

m  —  1       2 


•     ••..•     (^^^ 


•     • 


(Art.  46) 


.     (3) 


which  determines  x  -j-  ^a,  and  therefore  x. 

If  we  introduce  an  auxiliary  angle,  the  calculation  of 
equation  (3)  is  facilitated. 


n'iff?. 


132 


PLANE  TEIGONOMETRY. 


Thus,  let  m  =  tan<^ ;  then  we  have  by  [(14)  of  Art.  61] 

m  +  1      tan  (6  +  1         i.  /  .       t<-o\ 
— T--  =  - — ^-3—  =  cot  («^  -  4o  ), 

m  —  1      tan  <^  —  1 
which  in  (3)  gives 

tan  [a;  +  "]«=  cot  (</>  —  45°)  tan^a 

This,  with  tan  <^  =  wi, 

gives  the  logarithmic  solution. 

The  logarithmic  solution  of  the  equation 
sin  (a  —  a;)  =  m  sin  x 
is  found  in  the  same  manner  to  be 

tan  <^  =  m, 

and  tan  [  a;  —  ^  j  =  cot  (<^  -f  45°)  tan  -> 

which  the  student  may  show. 

Example.  —  Solve 

sin  (106°+  x)=z- 1.263  sin  a;  (a;  <  180°). 

log  tan  <^  =  logm  =  log  (  -  1.263)  =  0.10140  -. 

.-.  </,  =  128°  22'.3. 
«^-45°=    83°22'.3;      log  cot  (<^  -  45°)  =   9.06523 
ia=    53°    O'.O,  logtan^«  =  10.12289 


(4) 


logtan/'a;  +  -^=    9.18812 


a 


a;  +  -  =  8°46'.0  or  188°  46'. 

2 

.-.  a;  =  -44°  14'  or  135°  46'. 


L 


86.  Solve  the  equation 

tan  («  +  .t)  =  m  tan  X (1) 

in  which  a  and  m  are  given. 


TRIGONOMETRIC  EQUATIONS. 


133 


From  (1)  we  have 

tan  (u  -f  x)  -f-  tan  x  _m-\-l 
tan  (a  +  a;)  —  tan  x     m  —  1 

sin(a+  2x) 


sin  a 


.'.  sin(a -f-2a;)  =  — !— rSina 


[(21)  of  Art.  61] 
.    .    .    .    .    (2) 


where 


w  —  1 
=  cot(<^  -  45°)  sin  a  .    (Art.  85) 

tan  <f>  =  m. 


Example.  —  Solve  tan (23°  16'+  x)  =  .296  tana;, 
log  tan  <^  =  logm  =  log(.296)  =  1.47129. 
.-.  </)  =  16°  29'.3. 
,/,_45°=-28°30'.7;      log  cot(</) -45°)=  10.26502- 
«  =  23°16'.l,  log  sin  «=    9.59661 

log  sin(«  +  2i»)  =    9.86163- 
a-\-2x  =  220°  38'.::  or  313°  21'.1. 
.-.  a;=101°41'.5  or  145°2'.6. 


87.  Solve  the  equation 

tan  («  +  x)  tan  x  =  m 
in  which  a  and  m  are  given. 

From  (1)  we  have 

1  +  tan(«ie  +  x)  tan^  _  1  -fm 
1  —  tan  (rt  +  a;)  tan  x      1  —  m 

cos« 


(1) 


>■■'■■».       '■> .. 


cos(«  -\-'2x) 


(Ex.  4  of  Art.  47) 


.'.  cos(rt  +  2  a;)  = 


m 


cos« 


where 


1  -f-  m 

=  tan  (45°  -  <^)  cos  «    [  ( 16)  of  Art.  61] 
tan  ^  =  m. 


i.!i 


f,  t 


i 


134  PLANE  TESONOMETBY. 

Example.  —  Solve  tan (65°  +  a;)  tana;  =  1.519G(«  <  180°). 

log  tan  tf>  =  login  =  0.18173. 
.-.  «^  =  r.0='39'9"; 
45°  -  <^  =  -  11°  39'  9" ;       log  tan  (45° -<^)  =  9.31434- 
«  =  65°  0'  0" ;  log  cos  a  =  9.62595 

log  cos  (a  +  2a;)=  8.94029- 
«  + 2a;  =  95°  or  265°. 

.     2a;  =  30°  or  200°. 

.-.     X  =  15°  or  100°. 

88.  Solve  the  equations 

m sin  {B -\-x)=a (1)  . 

m  sin  (<^  4-  a;)  =  6 (2) 

for  m  and  x,  the  other  four  quantities,  6,  4>,  a,  h,  being 
known. 

Expanding  (1)  and  (2)  by  (Art.  44),  we  get 

msin^cosa +  mcos  ^sin  a;  =  a (3) 

wrin<^cosa;-f- mcos  <^  sin  a;=  6 (4) 

Multiplying  (3)  by  sin  «/!>  and  (4)  by  sin0,  and  subtracting 
the  latter  from  the  former,  we  liave 

m  sin  X  (sin  <^  cos  6  —  cos  <^  hin  6)  =  a  sin  <^  —  6  sin  ^. 

a  sin  <f>  —  6  sin  6  ,^. 

sin((^-^)  ^ 

To  find  the  value  of  mcos.i;,  multiply  (3)  and  (4)  by 
cos  <^  and  cos  tf,  respectively,  and  subtract  the  former  from 
the  latter.     Thus 

m  cos  x(sin  <^  cos  0  —  cos  <^  sin  0)=h  cos  0  —  a  cos  <^. 

ftcostf  — acosd*  ...^ 

.-.  mcosa;  = ^ ^ (6) 

sin(<^-e)  ^ 

Having  obtained  the  values  of  wsinx*  and  mcosa;  from 
(5)  and  (6),  in  and  x  can  be  calculated  by  Art.  83. 


TRIGONOMETlilC  EQUATIONS.  135 

EXAMPLES. 

1.  Solve   m  COS  (6  +  x)  =  a,   and  w  sin  ( </>  +  .r )  =  &,    for 

m  sin  X  and  w  cos  x.  a  ■  l>  t'<js  6  —  a  sin  c6 

^ns.  msin.'c=  ^, 

<v«„^o^      6sin^  + acos</» 
meosa;=    -         '  'r. 

2.  Solve  7n  cos  (^  +  cc)  =  a,  and  m  cos  (<^  —  a;)  =  b,  for 
m  sin  iB  and  m  cos  a;.  ^^^^   ^^  ^.^^  ^.  ^  ^/ cos  ^  -  «  cos  <^ 

sin(^  +  <^)     ' 

-.V,  „^„         6sin  ^4-rt  sin  <i 
7n  cos  X  =  —    — ^• 

sin(e  +  <^) 

89.  Solve  the  equation 

xcos  a -\-y  sin  H  =  m     .......     (1) 

xsiua  —  ?/cos«  =  w (2) 

for  X  and  y. 

Multiplying  (1)  by  cos  a  and  (2)  by  sin«,  and  adding, 

we  get 

x  =  m  cos  «  4-  w  sin  rt. 

To  find  the  value  of  y,  multiply  (1)  by  sina  and  (2)  by 
cos«,  and  subtract  the  latter  from  the  former.     Thus 

y  =  m  sin  a  —  n  cos  a. 

Example.  —  Solve 

a  sin  «  +  y  cos  «  =  a, 
x  cos  a  —  y  sin  a  =  h. 

90.  To   adapt  Formulte  to  Logarithmic  Computation. — 

As  calculations  are  performed  principally  by  means  of 
logarithms,  and  as  we  are  not  able  by  logarithms  directly* 
to  add  and  subtracit  (quantities,  it  becomes  necessary  to 
know  how  to  transform  sums  and  differences  into  products 

*  Addition  and  Siibtraelion  Tables  are  publislied,  by  means  of  which  the  logarithm 
of  the  sum  or  difference  of  two  numbers  may  be  obtained.  (See  Tufeln  der  Addi- 
tions, und  Siibtractious,  Logarilhineii  flir  Hiebcn  Stellen,  von  J.  Zecb,  Berlin.) 


1 1,  ?■■>■■-■' 


186 


PLANE  TRIGONOMETRY. 


and  quotients.  An  expression  in  the  form  of  a  jn'oduct  or 
quotient  is  said  to  be  adapted  to  logarithmic  coftipntation. 

An  angle,  introduced  into  an  expression  in  order  to  adapt 
it  to  logarithmic  computation,  is  called  a  Subsidiary  Angle. 
Such  an  angle  was  introduced  into  each  of  the  Arts.  84,  85, 
86,  and  87. 

The  following  are  further  examples  of  the  use  of  sub- 
sidiary angles : 

1.  Transform  a  cos  6  ±b  sin  6  into  a  product,  so  as  to 
adapt  it  to  logarithmic  computation. 

tan  0 ;  *  thus 

acos^±  &sin^  =  af  cos6  ± -sin0)    . 
=  a  (cos  6  ±  tan  <f>  sin  0) 


Put  ^ 
a 


a 


GOS<j> 


cos(^T  <^). 


a 


sin  (^±  <^). 


if 


Similarly, 

a  sin  6  ±b  cos  6  = 

QOStfi 

Transform  a  ±b  into  a  product, 

a  -f  6  =  a  ( 1  -f-     J  =  «  (1  +  tan-  <{>)  ~  a  sec^  </>, 

-  =  tan2<i. 
a 


a  —  6  =  a[l  —  )  =  a  cos^  <f>, 


if 


-  =  surdi. 
a 


*  The  fundaraental  foimulro  coB(jr  t  y)  and  sin  (a*  +  y)  (Art.  42)  afford  o::amp1eg  of 
one  term  equal  to  the  Bum  or  difference  of  two  terms ;  hence  we  may  transform  an 
expression  acosO  t  bsinO  Into  an  equivalent  product,  hy  conforming  it  to  the  for- 
muia*  Just  mentioned. 

Thus,  comi)aring  the  identity,  vi  cos  <fr  cos  9  +  m  sin  </>  sin  fl  =  m  cos  (0  ?  9)  or  m  cos 
(6  ^^),  witlincos0  t  bH\n0,  we  will  haven  cos  9  h  bfiine  =  mcoa(0  f  if>)  if  we  assume 


a-m  cos  (/>  and  6  = 
See  Art.  84. 


wising;  i.e.  (Art.  83),  if  tan</)  =  "  and  m  =  -^i— 

a  co»<f> 


8in<^i 


-  as  above. 


ADAPTATION  TO  LOGARITHMIC  COMPUTATION.    137 

.-.  log(a +  ?>)  =  loga +21ogsec<^; 
and  log  (a  —  6)  =  log  a  +  2  log  cos  <f}. 

4.    Transform  1239.3  sin  6  -  724.G  cos^  to  a  product. 

logft  =  log  (- 724.G)  =  2.8G010- 
log  a  =  log  (1239.3)  =  3.09318 

log  tan  (^  =  9.76692 - 

.-.  ,^  =  - 30°  18'.8.       ■         •  •' 

M 

loga  =  3.09318 
log  cos  <^  =  9.93615 

log-^  =  3.15703  ^    , 

cos<^ 

...  _^==  1435.6. 

COS<^  fl 

.-.  1239.3  sin  ^-724.6  COS  ^=1435.6  sin  (0- 30°  18'.8). 

f 

-    "       ■'  ,■  ■ 

91.   Solve  the  equations 

rcos<f>cosO  =  a (1) 

rcos^sin^  =  & (2) 

rsin<f}  =  c (3) 

for  r,  <f>,  and  $. 

Dividing  (2)  by  (1),  we  have 

tan^  =  -, 
a 

from  which  we  obtain  $. 

''    -..-'...■.   ^  ' 

From  (1)  and  (2)  we  have  . 

rcos^  =  -^  =  -;^- (4) 

cos^      sin^ 

from  which  we  obtain  rcos<^. 

From  (3)  and  (4)  we  obtain  r  and  <^  (Art.  83). 


1 
1^- 

1 

i 

'  * 

138 


PLANE  TRIGONOMETRY. 


1.   Solve 


for  r,  (f),  6. 


EXAMPLES. 

r  COS  </>  COS  6  —  —  53.953, 
rcos<^  sine  =  197.207, 
rsin</)  =  - 39.062, 


.'.  <^  =  -10°49'. 

logrt'os<^  =  2.31060 
log  cos  <^  =  9.99221 

logr=  2.31839 

.-.  r  =  208.16. 


log  &  =  2.29493 
loga  =  1.73201- 

log  tan  e  =  0.56292- 

.-.  e  =  105°18'.0. 
log  sing  =  9.98433 

log  r  cos  </)  =  2. 31060 
log  r  sin  <^  =  1.59175 - 

log  tan  <^  =  9.28115- 


92.  Trigonometric  Elimination.  —  Several  simultaneous 
equations  may  be  given,  as  in  Algebra,  by  the  combination 
of  which  certain  quantities  may  be  eliminated,  and  a  result 
obtained  involving  the  remaining  quantities. 

Trigonometric  elimination  occurs  chiefly  in  the  applica- 
tion of  Trigonometry  to  the  higher  branches  of  Mathematics, 
as,  for  example,  in  Physical  Astronomy,  Mechanics,  Analytic 
Geometry,  etc.  As  no  special  rules  can  be  given,  we  illus- 
trate the  process  by  a  few  examples. 

EXAMPLES. 

1.   Eliminate  <^  from  the  equations 

x=a cos  (f),  y  =  b sin  <f}. 
From  the  given  equations  we  have 


-  =  cos  <^,  ^  =  sin  <ft, 
a  b 


which  in 
gives 


cos'^*^  +  sin'^*^  =  1, 
1. 


TRIGONOMETRIC  ELIMINATIONS.  139 

2.  Eliminate  0  from  the  equations 

"•  a  cos  (}>-{- 1)  sin  <fi  =  c, 
b  cos  <f>  -{-c  sin  <f>  =  a. 

Solving  these  equations  for  sin  <^  and  cos  <^,  we  have 

.     .      be  —  a^ 
sm<^= , 

0^  —  ac 

,       c- —  ab 
coS(f>  = -; 

ac  —  b^ 
which  in  cos''  <^  +  sin^  <^  =  1, 

gives       {be  —  a^y  +  (c^  —  ab)''^  =  {ac  —  Wy. 

3.  Eliminate  <^  from  the  equations 

ycos<f)  — -  icsin  <^  =  acos2<^, 
ysm<f>  -\-xcos<f>  =  2  a  sin  2  <^. 

Solve  for  a;  and  y,  then  add  and  subtract,  and  we  get 
X  +  y  =  «(sin<^  +cos<^)(l  +  sin  2^), 
x  —  y  =  a(sin<^  —  cos  <fi)  (1  —  sin2<^). 

...   {x-{-yy  =  a\l  +  sm2<f>y, 
(x-yy  =  a\l-8in2cl>y. 

...  (x  +  yy  +  {x-y)^  =  2J. 

4.  Eliminate  a  and  /8  from  the  equations 

a  =  sin«cosj3sin  6  +  cosacostf      .     .     .  (1) 

6  =  sinrt  cos/8cos0  — cos«  sin^      .     .     .  (2) 

c  =  sinrt  sin/3  sin^ (3) 

Squaring  (1)  and  (2),  and  adding,  we  get 

a*  +  6''  =  sin^'rt  cos'^/SH- cos'^rt (4) 

-^  =  sm'us\n'l3 (5) 


I 

140 


PLANE  TRIGONOMETRY, 


Adding  (4)  and  (5),  we  have 

a- +  t' +.''■'    =1. 
sin=^^ 

6.   Eliminate  ^  from  the  equations 
a  sin  ^  +  h  cos  ^  =  c, 
a  COS  ^  —  6  sin  ^  =  d.       Ans.  a?  -\-W  ^c^  -{■  d^. 

6.  Eliminate  0  from  the  equations 

m  =  cosec  6  —  sin  tf, 

?i  =  sec  ^  —  cos  $.  m^n^  (m)  +  n*)  =  1. 

7.  Eliminate  $  and  <^  from  the  equations 

sin  6  +  sin  tf>  =  a, 
cos  0  -\-  cos  <f>  =  b, 
Gos{0-<l>)  =  c.  a2  +  62_2c  =  2. 

8.  Eliminate  a;  and  y  from  the  equations 

tan  X  +  tan  ?/  =  a, 
cot  a;  +  cot  y  =  b, 
x  +  y  =  c.  cot  c  = 

9.  Eliminate  <f>  from  the  equations 

X  =  cos  2  <^  +  cos  tji, 
y  =  sin2^  +  sin  <^. 

Ans.  2x=(x'^  +  fy-3(x^-[-y^). 


EXAMPLES. 

Solve  the  following  equations : 

1.  tan  d  + cot  ^  =  2. 

2.  2sin2^+V2cos^  =  2. 

3.  3tan2^-4sin-^  =  l. 

4.  2sin2(9  4-V2sin^  =  2. 


a 


1 

—  • 

b 


Ans.  45". 

90%  or  45°. 

45^ 

45". 


EXAMPLES. 


141 


5.  cos2^-V3  cos  ^4-^  =  0. 

6.  sin  5  ^  =  16  sin* ^. 

7.  sin9  6^  — sin^=sin4tf. 

8.  2  sin  ^  =  tan  0. 

9.  6cot20-4cos20  =  l. 

10.  tan  6  +  tan  (^  -  45°)  =  2. 

11.  cos^  +  V3sin^  =  V2. 

12.  tan  (^  +  45°)  =  1  +  sin  2  ^.. 

13.  (cot0-tan0)''(2  +  V3)  =  4(2-V3) 

14.  cosec  e  cot  ^  =  2  V3. 

15.  cosec  $  +  cot  0  =  V3. 

16.  sin- =  cosec  ^  — cot  ^. 

17.  sin  5^  cos  3^  =  sin  9^  COS  7^. 

18.  8in2^  +  cos2  26l  =  f. 

V3sin^-cos^  =  V2. 


Ans.  30" 


nrr,  or  nir  ±  - 
'  6 

iwTT,  or  |j.7r  ±  -^ 

it) 


nTT,  or  2  htt  i:  ; 


2.w±| 

iiTT  —  -,  or  nir 
4 

2n7r±^ 
o 

2n7r  +  ^7r, 

2M7r, 

^ig.n7r  +  (-l)- 


nTT  ±  :^,  or  7i7r  ± /^tt 


19. 

20.   tan  ^  + cot  ^  =  4 

21. 

22. 


IT 


TT 


V3 


V3 


sin  (^  +  «^)  = -^,  cos  (0  -  <^)  - 


4'  ^     12 


Solve  m  sin  ^j!.  «=  1.29743,  and  m  cos  <^  =  6.0024. 

Ans.  (<^<180°) 


142 


PLANE  TRIGONOMETRY. 


'fi 


23.  Solve  m  sin  «^=  -0.3076258,  and  m  cos  0=0.4278735. 
(m  positive.)  Ans.  ^  =  324°  17'  6".6,  m  =  0.52098. . 

24.  Solve  m  sin  <^  =  0.08219,  and  m  cos  </>  =  0.1288. 

25.  Solve  m  sin  </>  =  194.683,  and  m  cos  </>  =  8460.7. 

26.  If  a  sin  ^4-6  cos  d  =  c,   and   a  cos  ^  -j-  &  sin  d  =  c  sin  6* 
cos  6,  show  that  sin  2  0{c^  —  a-  —  ft'*)  =  2  a6. 

Solve  the  following  equations  : 


27.  V2sin^4-V2cos^  =  V3.  Ans.  _^-}-n7r +  (- l)"!"- 

28.  2  sin  a;  +  5  cos  a;  =  2.     Sug.  [2.5  =  tan  68°  12']. 

.4715.  a;  =  -  68°  12'  +  n  180°  +  ( -  1 )" (21°  48'). 

29.  3cosa;-8sina;  =  3.    *Sug.  [2.6  =  tan  69°  26' 30"]. 

Ans.  x  =  -  69°  26'  30"  +  2  m  180°  ±  (69°  26'  30"). 

30.  4  sin  X  -  15  cos  x  =  4.     Sug.  [3.75  =  tan  75°  4']. 

Ans.  X  =  75° 4'  +  n  180°  +  (-  1)"(14° 56'). 

31.  cos  («  4-  a;)  =  sin  («  +  x)  -\-  V2  cos  p. 

Ans.  a;  =  — «  —  ^4- Tnir  ± /?. 

32.  cos  ^  +  cos  3  ^  +  cos  5^  =  0. 

Ans.  |(2ri4-l)7r,  or  ^(37i±l)7r. 

33.  sin  5  ^  =  sin  3  ^  4-  sin  ^  =  3  —  4  sin^  6. 

Ans.  Hit  ±  ^,  or  ^(2  71  4-  1)^. 
o 


nir 


34.  2sin23'e4-sin26^  =  2. 

Ans.  H2n  +  l)7r,  or  :^  +  (- 1)"-^. 

35.  a(cos2^-l)4-2  6(cos(9  4-l)  =  0. 

Ans.  (2  7i4-l)7r,  or  cos~^-^^^^^ — 


a 


EXAMPLES. 


143 


36.  Solve 

m  sin(^  +  a;)  =  a  cos  /8,  and  m  cos(0  —  x)  =  a  sin  /?, 
for  msinic  and  mcosa;.     (Art.  67.) 

Ans.  m  sm  x  = ^^-— — -, 

cos  2  0 

m  cos  X  = ^^ '-' 

cos  U  0 

37.  Solve  mcos(6'  +  <^)  =  3.79,  and  m  cos  (^ -</>)  =  2.00, 
for  m  and  ^,  when  <f>  =  3r2TA.     (Art.  67.) 

38.  Solve  r  cos  <^  cos  ^  =  1.271, 

r  cos  <^  sin  6  =  —  0.981, 
r  sin  </.  =  0.890, 
for  r,  «^,  ^.     (Art.  70.) 


39.  Solve 

for  r,  «^,  d. 

40.  Solve 


r  cos  <^  cos  6  =  —  2, 

r  cos  (^  sin  6  =  -\-3, 

r  sin  <^  =  —  4, 

r  sin  «^  sin  ^  =  19.765, 
?•  sin  <f>  cos  6  =  —  7.192, 
r  cos  <^  =  12.124, 


for  r,  ift,  6. 

41.  Solve  cos (2 a;  4- 3 2/)  =  I,  cos(3a; +  2//)  =  ^V3. 

Ans.  X=^mr  ±^jrir  ±^'ir,  y  =  ^  nir  ±  ^ir  ±  ^ir. 

42.  Solve  cos3tf4-cos5e4-V2(cos^  +  3intf)cos^  =  0. 

Ans.  A$±$  =  2mr±lTr,  or  |(27i  +  l)7r. 

43.  Solve  cos3tf  +  sin3^  =  cose  +  sin^. 

Ans.  sin  ^  =  0,  or  tan  ^  =  —  1  ±  ■y/2. 

44.  Solve  3  sin  ^  4-  cos  ^  =  2  x,  sin  ^  +  2  cos  ^  =  a. 

Aris.  e  =  7rW,  a;=|Vl0. 


Mi  {'I 


^^'. 


»  !< 


i 


I'm  <:! 


144 


PLANE  TRIGONOMETRY, 


li 


45.  Solve  l.L'dS  sin  <>  =  0.1)48  +  m  sin (25°  27'.L'), 

1.2G8  cos  (^  =  0.281  +  m  cos (25°  27'.2). 

Ans.  <l>  =  00°  53'.8,  m  =  0.872. 

46.  Transform   x* -\- y*  +  z*  - 'J  fz- - 'J  z'x- - '2  x'lf  into  a 
product.       Ans.  —{x-\-y-\-z){y-{-z  —  x){z+x—ij){x-\-y—z). 

47.  Eliminate  6  from  the  equations 

m  sin  20=  n  sin  ^,  j:>  cos  26  =  q  cos  ^. 

^n«.  m^  +/)*  =  n^  -<-  ^^ 

48.  Eliminate  6  and  <^  from  the  equations 

x=:a  cos"*  ^  cos"  tfi,  y=b  cos"*  tf  sin"  <f>,  z  =  c  sin"  ^. 

A71S,  f  -  l^H-  ■,"»+      )"=l. 
\aj       \bj       \cj 

49.  Eliminate  $  from  the  e<iUiitions 

a  sin  ^  +  6  cos  B—.h,  a  cos  ^  —  6  sin  ^  =  k. 

Ans.  (t*  +  Z;-  =  h^  4-  fc^ 

60.  Eliminate  6  from  the  equations 

a  tan  6  -\-b  sec  ^  =  c,  a'  cot  0  -{-  b'  cosec  d  =  c'. 

^715.  (a'6  +  cb'y  +  («6'  +  c'by  =  (cc'  -  aa')='. 

61.  Eliminate  $  from  the  equations 
a  =  2 a  cos ^ cos 2$  —  a  cos d, 

y  =  2a  cos  tf  sin  2  ^  —  a  sin  ^.  ^ns.  0^4-3/'  =  «^ 

62.  Eliminate  ^  from  the  equations 

x  =  a  cos  ^  +  6  cos  2  d,  and  2/  =  ti  sin  0  -\-b  sin  2  ^. 

Ans.  a'l{x  +  by  +  y'}  =  [x"  +  y' - 6']'. 

63.  Eliminate  «  and  /3  from  the  equations 
6  +  ccosa  =  wcos  («  —  tf), 

6  4-ccos/8  =  ttcos  (/3  — ^),  a  —  (3  =  2<l>; 
and  show  that  " 

M*  —  2  Mc  cos  ^  +  c^  =:  &*  sec''  ^. 


EXAMPLES  IN  ELIMINATION. 


145 


54.    Eliiniiiato  6  ami  «/>  from  tlie  e(|iuitiuns 

xcosO  +  y8m  6  =  a,  bain  {6  +  «^)  =  asiii  </>, 
xoos{0-{-2<i>)  —  y  sin  {$  +  2  <^)  =  a. 


nY 


55.   Eliminate  0  from  the  e(j[uations 
a     sec''^ +  cos'*d' 


y 


=  sec^^  +  cos^^. 


56.   Eliminate  6  from  the  equations 

(a  +  6)  tan  (^  —  <^)  =  (a  —  6)  tan  (^  4- <^), 

67.    Eliminate  0  from  the  equations 

•    a  A        fzr~, — 2    cos^^  ,  sin'-'^  1 

or  Ir        x^  +  7f 

Ans.  ^+-^  =  1. 
a^     b^ 

58.  Eliminate  6  and  ^  from  the  equations 

a^  cos^  6  —  b"^  cop^  f{>  =  c^,  a  cos  6  -\-b  cos  4>  —  r, 
a  tan  ^  =  ?;  tan  <f>. 

59.  Eliminate  ^  from  the  equations 

»isin^  — ?rtcos^  =  2msin<^,  , 

n  sin  2$  —  m  cos  2  <^  =  n. 

^4ns.   (wsin^  +  mcos0)^  =  2?n,(m +  7i). 

60.  Eliminate  a  from  the  equations 
a;  tan  («  — /8)  =  2/ tan  (rt  4- /8), 

(a;  —  y)  cos  2  «  -f  (a;  +  y)  cos  2  /8  =  2. 

^7}s.  «^  +  4 a;y  =  22  (ic  4-  2/)  cos 2)3. 


II 


t 


it, 


W.    '•"' 


!ii^ 


146 


PLANE  TBIGONOMETRY, 


CHAPTER  VI. 

BELATIONS    BETWEEN    THE   SIDES    OF  A  TKIANGLE 
AND  THE  FUNCTIONS  OF  ITS  ANGLES. 

93.  FormulsB.  —  In  this  chapter  we  shall  deduce  forinulye 
which  express  certain  relations  between  the  sides  of  a  tri 
angle  and  the  functions  of  its  angles.  These  relations  will 
be  applied  in  the  next  chapter  to  the  solution  of  trianglen. 
One  of  the  principal  objects  of  Trigonometry,  as  its  name 
implies  (Art.  1).  is  to  establish  certain  relations  between 
the  sides  and  angles  of  triangles,  so  that  when  some  of 
these  are  known  the  rest  may  be  determined. 

RIGHT  TRIANGLES. 

94.  Let  ABC  be  a  triangle,  right-angled  at  C.  Denote 
the  angles  of  the  triangle  by  the  let- 
ters A,  B,  C,  and  the  lengths  of  the 
sides  respectively  opposite  these  an- 
gles, by  the  letters  a,  b,  c*  Then  we 
have  (Art.  14)  the  following  relations :    /^ 


o  =  csin A  =  ccosB  =  6tan  *  =  6cotB  . 
b  —c  sin  B  =  c  cos  A  =  a  tan  B  =  <t  cot  A  . 
c  =  6  sec  A  =  asecB  =  ftcosec  B  =  a  cosec  A 


(1) 

(2) 


T/hich  may  be  expressed  in  the  following  general  theorems : 

*  The  itndent  mint  remember  that  a,  b,  c,  nre  numhern  czpreiiainK  the  lengthH  of 
the  iides  In  terms  (  some  unit  of  length,  such  an  a  foot  or  a  mile.  The  unit  may  be 
whatever  we  please,  but  muat  be  the  aame  for  all  the  aides. 


OBLIQUE  TRIANGLES. 


147 


I.  In  a  right  triangle  each  side  is  equal  to  (he  product  of  the 
hypotenuse  into  the  sine  of  the  opposite  angle  or  the  cosine  of 
the  adjacent  angle. 

II.  In  a  right  triangle  each  side  is  equal  to  the  product  of 
the  other  side  into  the  tangent  of  the  angle  adjacent  to  that 
other  side,  or  the  cotangent  of  the  angle  adjacent  to  itself 

III.  In  a  right  triangle  the  hypotenuse  is  equal  to  the 
product  of  a  side  into  the  secant  of  its  adjacent  angle,  or  the 
cosec  mt  of  its  opposite  angle. 

EXAMPLES. 

In  a  right  triangle  ABC,  in  which  C  is  a  right  angle, 
prove  the  following : 


1.    tan  B  =  cot  A  -f  cos  C. 
3.   cos2A-i/^.os2B  =  0. 

6.   cosec  21  (=  ^  +  — 


2.   sin2A  =  sin2B. 
2  ah 


7.   tan  2  A 


26  ■  2a 

'2ah 
*-a^' 


4.  sin  2  A  = 
6.  co82A  = 
8.   sin3A  = 


'Aab''-a^ 
c3 


ex 


OBLIQUE  TRIANGLES. 

95.  Law  01   Sines.  —  In  any  triangle  the  sides  are  pro- 
portional to  the  sines  of  the  opj)osite  angles. 

Let  ABC  be  any  triangle.  Draw 
CI)  perpendicular  to  AB. 

We  have,  then,  in  both  figures 

CD-easinB  =  6sinA.     (Art.  94) 
.'.  asinB  =  6  sin  A.  ^ 

•  _?_  — _A_. 

sin  A     sin  B 

Similarly,  by  drawing  a  perpen- 
dicular from  A  or  B  to  the  opposite 
side,  we  may  prove  that  A 


148 


PLANE  TRIGONOMETRY. 


or 


ft             c 

and 

sin  V>     sin  C' 

* 

a            b 

c 

a 


sin  C     sin  A 


sin  A      sin  B      sin  C 
a:b:c  =  sin  A  :  sin  B  :  sin  C. 


96.  Law  of  Cosines.  —  In  any  triangle  the  square  of  any 
side  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides 
minus  twice  the  product  of  these  sides  and  the  cosine  of  the 
included  angle. 

In  an  acute-angled  triangle  (see 
first  figure)  we  have  (Geom.,  Book 
III.,  Prop.  26) 


•on' A  ri2 


BC  =  AC  +  AB  -  2  AB  X  AD, 
or     0=*  =  62-1- c'- 2c.  AD. 
But  AD  =  6  cos  A. 

.'.  a'  =  5'  +  (r*  —  26c cos  A. 

In  an  obtuse-angled  triangle  (see 
second  figure)  we  have  (Geom., 
Book  III.,  Prop.  27) 


n2 


or 


But 


BC  =  AC  +  AB'  +  2  AB  X  AD, 

a^  =  b^-{-c'  +  2c-AI). 
AD  s=  6  cos  CAD  =  —  b  cos  A. 
.'.  a*  =  b^-\-(?  — 2  be  coa  A. 
Similarly,      6*  =  c*  -f-  a*  —  2  ca  cos  B, 
c'  =  o'  +  6'-2a6cosC. 


Not*.  —  When  one  equnMon  in  the  solution  of  triangleii  has  been  obtained,  the 
other  two  may  generally  be  obtained  by  advancing  the  letters  so  tbat  a  becomes  h, 
b  becomes  r,  and  c  becomes  n;  the  order  is  abc,  bca,  nib.  It  Is  obvious  that  the 
forroule  thus  obtained  are  true,  since  the  naming  of  the  sides  makes  no  difference, 
provided  the  right  order  is  maintained. 


OBLIQUE  TRIANGLES.  149 

97.  Law  of  Tangents.  —  In  any  triangle  the  sum  of  any 
two  sides  is  to  their  difference  as  the  tangent  of  half  the  sum 
of  the  opposite  angles  is  to  the  tangent  of  half  tfieir  difference. 

By  Art.  95,     a  :  6  =  sin  A  :  sinB. 

By  composition  and  division, 

a  +  h  _  sin  A  +  sin  B 

a  —  h      sin  A  —  sin  B         ,      ■ 

Similarly.      ^+-«  =  *5tEi(liJlC) 

^       b-c     tan^(B-C)  .:     ^  ^ 

c  +  a_tan^(C  +  A)  .^. 

•'     c-a      tan^(C-A) ^  ^ 

Since  tan^(A +  B)  =  tan(9()°  ~iC)=  cot^C, 

the  result  in  (1)  may  be  written 

a  +  &  _       cot|C  .^x 

a-b     tan^(A-B) '     ^  - 

and  similar  expressions  for  (2)  and  (3). 

98.  To  show  that  in  any  triangle  c  =  a  cos  B  -f  &  cos  A. 

In  an  acute-angled  triangle  (first  figure  of  Art.  90)  we 
l^a.ve  c  =  DB  +  DA 

=  a  cos  B  +  6  cos  A. 
In  an  obtuse-angled  triangle  (second  figure  of  Art.  96) 
we  have  c  =  DB-DA 

=.:  a  cos  B  —  6  cos  CAD. 

.*.  c  =  acosB  +  6cos A. 

Similarly,  6  =  c  cos  A  -f-  a  cos  C, 

a  —  b  cos  C  -f-  c  cos  B. 


*->^ 


ill 


.1 


160 


PLANE  TRIGONOMETHY. 


EXAMPLES. 


1.  In  the  ti^iangle  ABC  prove  (1) 

a-\-h:c=  cos  |(A  —  II)  :  sin  \  C, 
and  (2)  a  —  6  :  c  =  sin  ^  ( A  —  B)  :  cos  |  C. 

2.  If  AD  bisects  the  angle  A  of  the  triangle  ABC,  prove 

BD :  DC  =  sin  C :  sin  B. 

3.  If  AD'  bisects  the  external  vertical  angle  A,  prove 

BD':CD'  =  sinC:sinB. 

4.  Hence  prove      1   ^ 2 cos  j A cosHB^zCj.  ' 


BC 


asinB 


and  also 


1    ^2sin^Asin^(C-B) 
D'C  asinB 


99.  To  express  the  Sine,  the  Cosine,  and  the  Tangent  of 
Half  an  Angle  of  a  Triangle  in  Terms  of  the  Sides. 

I.   By  Art.  96  we  have 


cos  A  = 


62  + c» 


a' 


Let 
then 


26c 


=  1  —  2  sin 


*  9     •^^ 

1  r»  *  _ 


2 


(Art.  49) 


2sin*A  =  i_^±^_?: 
2  26c 

^a8-(6-c)» 

^         26c 

—  (ct  +  6  —  c)  {a  - 
■"  26c 


i^ 


6tc), 


a  +  6  4-c  =  2«; 

a  +  6  — c  =  2(.9--c),  and  o  — 6 +  c  =  2(s  — 6). 
.,  28in'^  =  2i«-c)2(^-61. 
2  26c 


sin 


^v 


is-h)U-cX    .... 


6c 


(1) 


OBLIQUE  TUIANGLEH.  151 

Similarly,    sin|^  =  ^/5H3SEl0  .     .     .     .     .     ."    (O) 

II.  cosA  =  2cos2^-l (Art.  49) 

.'.  2cos^     =  1  H -^ — 

2  26c 

26c 

_  (g  +  6  +  c)  (6  +  c  —  g) 
26c 

2s.2(s-g) 

,    "     II-  ■  —  ■..-^ z.  » 

26c 
A         js(s  —  a)  /,. 

Similarly,    co3|  =  J'^'"**) •     •     (5) 

„^„C         /s(s  —  c)  ,^. 

cos— =  a/-^' ^ (6^ 

2       \      g6  ^^ 

III.  Dividing  (1)  by  (4),  we  get 

-t=V^^^!^ (^) 

Similarly/tan?  =  x/5H3IZE«l (8) 

tau5=J5^;SZES (9) 

2       \f       s{s-c)  ^  ^ 

Since  any  angle  of  a  triangle  is  <  180°,  the  half  angle  is 
<  90° ;  therefore  the  positive  sign  must  be  given  to  the 
radicals  which  occur  in  this  article. 


J'  *'  ■? 


'i 


152 


PLANE  TRIGONOMETRY. 


100.  To  express  the  Sine  of  an  Angle  in  Terms  of  the 
Sides. 


sin  A  =  2  sin  —  cos  — 


(Art.  49) 


\  be  M      be 


(Art.  99) 


/.  sin  A  =  -  V«  (s  —  a)  (s  —  b){s  —  c). 
be 


Similarly,   sin  B  =  -  Vs  (s  —  a)  (s  —  6)  (.s  —  c), 

ac 


2 


Cor. 


sin  C  =  —  Vs  (s  —  a)  (s  —  6)  (s  —  c). 
1   ,  /o^2 


sin  A  =  -!-  V"26V-f" 2cV+  2a262_  a*-  b*-  c\ 
2  be 


and  similar  expressions  for  sin  B,  sin  C. 


EXAMPLES. 

In  any  triangle  ABC  prove  the  following  statements : 

1.  a  (6cosC  —  ccosB)  =  6^^  — cl 

2.  (6  +  c)cosA-f(c  +  a)cosB  +  (a  +  6)cosC  =  a  +  6  4-c. 

o    sin  A  +  2sinB  _  sinC 
a +  26        ~"7~' 

4     sin' A  — m  sin'' B  _  sin'C 
a^-mb^        ~~~^' 

5.    a  cos  A  +  6  cos  B  —  c  cos  C  =  2  c  cos  A  cos  B. 


cos  A 


sin  B  sin  C 


+ 


cosB 


7.  a8in(B-C) 

8.  tan  ^  A  tan  ^ ; 


sin  C  sin  A 

+  b  sin  (C 

B-*-^. 

+  -.- 


cosC 


sin  A  sin  B 


^2. 


-A)  +  csin(A-B)  =  0. 


9.    tan^A-4-tan^B  =  (.s-6)H-(.'»~c). 


AREA  OF  A   TRIANGLE. 


153 


101.  Expressions  for  the  Area 
of  a  Triangle. 

(1)   Given  two  sides  and  their 
included  angle. 

Let  S  denote  the  area  of  the  tri- 
angle ABC.     Then  by  Geometry, 

2S  =  cxCD. 


But   in  either  figure,   by   Art, 


94, 


CD  =  Z>sinA. 
.-.  S  =  1 6c  sin  A. 
Similarly,         S  =  J[^  ac  sin  B, 
S  =  ^  a6  sin  C. 

(2)  Given  one  side  and  the  angles. 


Since 


which  is 


a'.b  =  sin  A  :  sin  B (Art.  95) 

,      a  sin  B 
sin  A 

S  =  ^  n6  sin  C,  gives 

a- sin  B  sin  C 


S  = 


Similarly, 


S  = 


2  sin  A 
h-  sin  A  sin  C      c^  sin  A  sin  B 


2sinB 


2  sin  C 


(3)   Given  the  three  sides. 


sin  A  =  ^  \/s{s  -  a)  (s  -  b)  (s  -  c)    (Art.  100) 
be 


Substituting  in 


we  get 


S  =  ^  6c  sin  A, 

S  =  \/s(s-a){s-b)(s-c). 


I 


154 


PLANE  TinGONOMETRY. 


102.  Inscribed  Circle.  —  To  find  the  radius  of  the  iusanbed 
circle  of  a  triamjle.  C 

Let  ABC  be  a  triangle,  0  the 
centre  of  the  inscribed  circle,  and 
r  its  radius.  Draw  radii  to  the 
points  of  contact  D,  E,  F ;  and  join 
OA,  OB,  OC.    Then  a 


c   D 


S  =  area  of  ABC 
=A  AOB  +  A  BOC  +  A  CO  A 

=  ^  re  4-  i  »*«  +  i  t'h 

a-\-h  -\-c 
—  r  —  =  rs     .     .     .     . 


(Art.  99) 


...  r  =  ^  =J('-^)(''-miJ^l     .   (Art.  101) 

S  \  8 

t 

103.  CircuniBcribed  Circle.  —  To  find  the  radius  of  the 

circumscribed  circle  of  a   triangle  in  ^q 

terms  of  the  sides  of  the  triangle. 

Let  0  be  the  centre  of  the  circle    Ai 
described   about  the  triangle   ABC, 
and  R  its  radius. 

Through  O  draw  the  diameter  CD 
and  join  BD.  D" 

Tken       Z  BDC  =  Z  BAC  =  Z  A. 


.-.  BC  =  2  R  sin  A,  or  «  =  2  R  sin  A. 
a  b  c 


But 


.-.  R  = 

sin  A  = 


2  sin  A     2  sin  B      2  sin  C 
6c 


Tj      ahc 


.    .     (1) 
(Art.  101) 

.     .     (2) 


flli 


BADII  OF  THE  ESCRIBED   CIRCLES. 


165 


104.  Escribed  Circle.  —  To  Jim! 
the  radii  of  the  escribed  circles  of  a 
trianyle. 

A  circle,  which  touches  one  side 
of  a  triangle  and  the  other  two 
sides  produced,  is  called  an  escribed 
circle  of  the  triangle. 

Let  O  be  the  centre  of  the 
escribed  circle  which  touches  the 
side  BC  and  the  other  sides  pro- 
duced,* at  the  points  D,  E,  and  F, 
respectively,  and  let  the  radius  of 
this  circle  be  rj. 

We  then  have  from  the  figure 

A  ABC  =  A  AOB  -f  A  AOC  -  A  BOC. 


S 


n  = 


8  — a 


r,(s-a).     (Art.  99) 
(1) 


Similarly  it  may  be  proved  that  if  rg,  r.j  are  the  radii  of 
the  circles  touching  AC  and  AB  respectively, 

S  S 


ro  = 


8-6 


r.= 


y    '3 


105.  To  find  the  Distance  be- 
tween the  Centres  of  the  Inscribed 
and  Circumscribed  Circles^  of  a 
Triangle. 

Let  I  and  0  be  the  incentre 
and  circumcentre,  respectively,  of 
the  triangle  ABC,  I A  and  IC 
bisect  the  angles  BAG  and  BCA ; 

*  Often  called  the  incentre  and  circumcen- 
tre of  a  triangle. 


166 


PLANE  TRIGONOMETRY, 


{1^  ; 


therefore  the  arc  BD  is  equal  to  the  arc  DC,  and  DOH 
bisects  BC  at  right  angles. 

Draw  IM  perpendicular  to  AC.     Then 


ZDIC 


_A  +  C_ 


BCD  +  BCI  =  DCI. 


Also, 


that  is, 


,  DI  =  DC  =  2Rsin^. 

A  A 

AI  =  IM  cosec  —  =  r  cosee  — • 

2  2 

.-.  DI.AI  =  2Rr  =  EI.IF; 
(R  +  0I)(R-0I)  =  2Rr. 

.-.  Ol'  =  R'^-2Rr. 

EXAMPLES. 


1.  The  sides  of  a  triangle  are  18,  24,  30 ;  find  the  radii 
of  its  inscribed,  escribed,  and  circumscribed  circles. 

Ans.  6,  12,  18,  36,  15. 

2.  Prove  that  the  area  of  the  triangle  ABC  is 


2cotAH-cotB 

3.  Find  the  area  of  the  triangle  ABC  when 

(1)  a  =  4,  &  =  10  ft.,  C  =  30*.  Ans.  10  sq.  ft. 

(2)  6=5,  c  =  20  inches,  A  =  60°.  43.3  sq.  in. 

(3)  a  =  13,  6  =  14,  c  =  15  chains.         84  sq.  chains. 

>!     T»  1       1       1       1 

4.  Prove  -  =  — H \-  — 

r     Ti     r-a     rg 


-    T.  a  sin  4  B  sin  4  C 

6.  Prove  r  = -—,  ^   ^    » 

cos^  A 


EXAMPLES. 


157 


6.  ove  that  the  area  of  the  triangle  ABC  is  represented 
by  each  of  the  three  expressions  : 

2 II*  sin  A  sin  B  sin  C, 

ra,  and 

Rr(sin  A  +  sin  B  -f-  sin  C). 

7.  If  A  =  60°,  a  =  V3,  b  =  V2,  prove  that  the  area 

-=i(3  +  V3). 

8.  Prove  R(sin  A +  8inB  +  sinC)«=  s. 

9.  Prove  that  the  bisectors  of  the  angles  A,  B,  C,  of  a 
triangle  are,  respectively,  equal  to 

A  B  C 

2  be  cos  —    2ca  cos  —    2  ab  cos  — 


o 


2 


6  +  c 


c  +  (t 


a  +  6 


106.  To  find  the  Area  of  a  Oyclic* 
Quadrilateral.  ^ 

Let  ABCD  be  the  quadrilateral,  and 
a,  b,  c,  and  d  its  sides.     Join  BD. 

Then,  area  of  figure  =  S 

=  I  ad  sin  A  +  ^  6c  sin  C 

=  ^(ad4-6c)sin  A    ...     (1) 

Now  in  A  ABD,    BD'  =  a'  +  d''-2  ad  cos  A, 

and  in  A  CBD,  B&  =b^  +  c^  ~2bc  cos  C 

—  ¥  -\-c^  —  2  be  cos  A. 

.:  cosA  = — — . 

2  {ad -\- be) 


\        L    2{ad-{-bc)     J 


y(2ad-f  2&c)'-(a'-  b'--c^  +  d^y 
2  {ad  +  be) 

*  See  Geometry,  Art.  251. 


I 


I 


m 


m 


168  PLANE  THIQONOMETHY, 

^  ^/[Ta  +  dy-  (b-  c)n  [(6  +  0)'-  («"^~rf?] 

2(ad  +  6c) 

y/(a  +  d  +  b  —  c)  ( a+'d -fe+c)  {b+c+a—d)(b + r^ci  4-  d ) 
■^  2{ad  +  bc) 

_2  V(<  -  g)  («  -  6)  (a  -c)(a-  d) 
ad  +  bc 

(where  2«  =  a -f- 6 -I- c  4-d). 

Substituting  in  (1),  we  have 

S  ^  V(«  -  a)  (a  -  b)  (s  -  c)  (s  -  d). 

The  more  important  formula  proved  in  this  chapter  are 
summed  up  Jis  follows : 

1.  ^^  =  T— 5  =  -T-p; (Art.  95) 

sin  A     sinB      smC 

2.  a»-6»  +  c»-26cco8A (Art.  96) 

3    ^L±6^t^i(A  +  B) (Art.  97) 

a-b     tani(A-B)  "^  / 

4.  sin^A=  J^^"^^^^~^^ (Art.  99) 

5.  cos^A  =  \r-^^^ — ^• 

^       be 

6.  taniA  =  \/^^^^SIEiI. 

^         \      8{8-a) 

7.  sin  A    =  ^  V«(s  -a)  {8-  b)  (s  -  c)    .     .  (Art.  100) 

be 

=::  _L  V2  6  V  +  2  c»a»  +  2  a'6»  -  a^  -  6^  -  c« . 
26c 

8.  Area  of  A  =  V«  («  -  a)  (a  -  6)  (s  -  c)  .     .  (Art.  101) 


EXAMPLES. 


169 


9.    Area  oi  A  =  Ua-\- b -\-c)=  rs    .     .     .     .  (Art.  102) 
10.  \ 


J.d  -a){»-h){i-c) 


11. 


K  = 


ahc 
4S 


(Art.  103) 


EXAMPLES. 


In  a  right  triangle  ABC,  in  which  C  is  the  right  angle, 
prove  the  following : 

\  n-n      sin^A  — sin*H 


J.* 

sur  A  +  sin^B 

.  oB     c  —  a 

2. 

sin''-= -• 

2        2c 

/       A  ,     .    AV     a  +  c 

3. 

(cos-  -f  sin  -  )  =—1-. 

\,       2            2)         c 

4. 

2  A      b  +  c 

COS''—  =  — ■ — ♦ 

3       2# 

6. 

sin(A-B)-f-cos2A  =  0. 

^a  —  b.A  —  B 

6.    =  tan • 

a  +  b  2 

7.  sin(A-B)4-8in(2A4-C)  =  0. 
a 


8.    tan  ^  A 


b-\-c 


9.    (sinA-sinB)'+(cosA4-cosB)'  =  2. 

IsinA 


10.    Jl±^-f  J?LzJ  =  -2^ 


+  6      Vcos2B 

In  any  triangle  ABC,  prove  the  following  statements 

11     /     I  I.N   •    C  A-B 

11.    (a4-o)sin     =cco8 — - — • 


i 


K 


liiM 


160 


PLANE  TRIGONOMETRY. 


12. 
13. 
14. 


(o  —  c)  cos  -  =  asm • 

^         ^        2  2 

a(6*+o')co8A+6(c''+a»)co8B+c(a*+6=)cosC=3a6c. 

a  —  ft  _  COS  B  —  cos  A 

c  1  +  cos  C 

6  4-  c     cos  li  +  cos  C 


16.  i_i:.'^  = 


a 


1  —  cos  A 


% 


16. 

17. 

18. 
19. 

20. 

21. 

22. 

23. 

24. 

25. 

26. 

27. 
28. 


Woe  sin  B  sin  C  = • 

ft  +  c 

ff -f-ft-f-c  =  (6  +  c)cos  A  -)-  (o  -f  a)cos  B  -|-  (a  -f  ft)eo8  C. 
6+c— a  =  (6  -f  c)cos  A  —  (c  —  a)co8B  -|-(a  —  ft)co8C. 
a  cos  ( A+  B  +  C)  -  ft  cos  (B  +  A)  -  c  cos  ( A  +  C)  =  0. 

cos  A  ,  cos  B  ,  cos  (y      a?  -f  ft'  -|-  c* 

H -, r 


a 

tan  A 


ft  c 

a  sinC 
ft  —  a  cos  C 


2aftc 


ft  cos' — h  c  cos'—  =  «. 
2  2 

.      B.      C     ft  +  c-o 
tan  -  tan  -  =  ~ 

2        2     ft+c+a 

A  J^ 

tan  -  (ft  +  c  —  a)  =  tan  -  (c  -f  a  —  ft). 

2  ^ 

c^  =  (ffl  -;-  ft)'sin'^  4-  (a  -  ft)'cos'^. 

c(co8A4-cosB)  =  2(rt  +  ft)v'n>^.  ' 

c 

c(co8  A  -~  cos  ^)  —  l^(ft  —  a) cos'-* 

tan  B  +  tan  C  =  (a'  4-  ft'  -  c')  -.  (a'  -  ft'  +  c^). 


EXAMl'LKS. 


161 


20.   rt«  -f.  ftn  4-  ,-2  =  2 (a&  cos  C  +  be  cos  A  -f-  ca  cos  B) . 

30.    C08»^  H-  cos^J  =  («  -  rt)  -f-  &(s  -  6). 

C  H 

,'U.   6  sin'—  +  c  sin*  -  =  .s  —  a. 

2  i 


32.    If  />  is  the  length  of  the  perpendicular  from  A  on 
BC, 


sin  A  =  -"• 
6c 


33.  If  A  =  3B,  then8inB  =  JJ^^--^. 

34.  It  v6t'  sin  B  sm  C  = j  ^-^^^  "  =  C* 

35.  acos^30s  ^^Josec-=:s. 

36.  If  cos  A  =  f,  and  cos  B  =  |jj,  then  cos  C  =  -  \\. 

37.  If  sin^B  +  sin='C  =  sin'A,  then  A  =  1M)°. 

38.  If  D  is  the  middle  point  of  BC',  prove  that 

4Al)'=26»4--^'-'-al 

39.  If  a  =  26,  and  A  =  3  B,  prove  that  C  =  60°. 

40.  If  1),  P:,  K,  are  the  middh;  points  of  the  sides,  BC,  CA, 
AB,  prove 

4(Al)'  4-  be'  +  CF')  =  3(r.»  +  6'^  +  c»). 

41.  If  a,  6,  c,  the  sides  of  a  triangle,  are  in  arithmetic 

progression,  prove 

,      A,      C      1  ,  . 

tan -tan  2  =  3. 

42.  If  taq  A  -  tan  B  ^  c  -  6^  ^^^^  ^  ^  g^o 

tan  A  +  tan  B         c 

43.  If  cos  L    -  '  .  prove  that  B  =  C. 

2sm{  . 


162 


PLANE  TRIGONOMETRY. 


41.    If  rt»  =  fc»  -  6c  4-  c",  prove  that  A  =  60°. 

46.  If  the  sides  of  a  triangle  are  a,  6,  and  Va*  +  ab  -\-  6-, 
prove  that  its  grejitest  angle  is  120°. 

4<).  I'rove  that  the  vertical  angle  of  any  triangle  is 
divided  by  the  median  which  bisects  the  base,  into  seg- 
ments whose  sines  are  inversely  proportional  to  the  adja- 
cent sides. 

47.  If  AD  be  the  median  that  bisects  BC,  j^rove  (1) 

(6«  -  c»)  tan  ADB  =  2  be  sin  A, 
and  (2)       cot  BAD  -f  cot  DAC  =  4  cot  A  +  cot  1^  -f  cot  C. 

48.  Find  the  area  of  the  triangle  ABC  when  a  =  (>25, 
6  =  505,  c  =  904  yards.  Ans.  151872  sq.  yards. 

49.  Find  the  radii  of  the  inscribed  and  each  of  the 
escribed  circles  of  the  triangle  ABC  when  (1  =  13,  6  =  14, 
c=16.  Ans.  4;  10.5;  12;  14. 

60.   Prove  the  aroa  S  =  ^  a'  sin  B  sin  C  coscc  A. 

61. 


62. 


« 


<( 


i( 


li 


« 


"  =  Vrrirjr,. 


«     « 


2o6c     /    „A„„B_     C 

[( — 


cos     cos  — cos 


} 


a-f  6-fcV       '^        '*        '* 

63.  Prove  that  the  lengths  of  the  sides  of  the  pedal  tri- 
angle, that  is,  the  triangle  formed  by  joining  the  feet  of  the 
perpendiculars,  are  a  cos  A,  b  cos  B,  c  cos  C,  respectively. 

64.  Prove  that  the  angles  of  the  pedal  triangle  are, 
respectively,  tt  —  2  A,  tt  —  2 1?,  tt  —  2  C. 

66.  Prove  r,rjr,  =  r'  cot*  -  cot"  -  cot'  -• 

2         2         2 

t^a     iJ  ABC 

c6.   Prove  r,  cos  -  =  acos— cos — 

•        2  2        2 

67.  Prove  that  the  area  of  the  incircle ;  area  of  the  tri- 
angle :  :  w  :  cot  —  cot   -  cot  -•  ,  -  > 

*  9       2       2 


EXAMPLES. 


163 


Prove  the  following  statements  : 

68.   If  a,  b,  c,  are  in  A.  1*.,  then  ac  —  6  rR. 

59.  If  the  altitude  of  an  isosceles  triangle  is  equal  to  the 
base,  R  is  five-eighths  of  the  base. 

60.  6c  =  4ir(cosA  +  coslUosC). 

61.  If  C  is  a  right  angle,  2r  +  2R  =  a-f6. 

62.  r^r^  +  rgr,  +  r,r2  =  5^. 

63.  l  +  l  +  i  =  _L. 
he     ca     ah     2rR 

C 

64.  r^-f-  r2  =  ccot -• 

65.  r  cos  -  =  a  sin  -  sin  -  • 

2  2       2 

66.  If  jp„  Pa,  Pn  be  the  distances  to  the  sides  from  the 
circumcentre,  then 

a  .h  .   c aha 


P\      P2     Pi      ^PiPiPa 
67.   The  radiuc  R  of  the  circuincircle 


—  ^   a/  abc 

~2\sinA8inB 


B  sin  C 


n* 


h^ 


68.   S  =  --  sin  2  B  4-  -  sin  2  A. 
4  4 


69. 


1  +-^.+  ^ 


1      4R 


s  —  a     a  —  b     8  ~  c     s        S 
7(».   a6rr  =  4R(«-a)(s-6)(.s-c). 

71.  The  distances  between  the  centres  of  the  inscribed 
and  escribed  circles  of  the  triangle  ABC  are  4  R  sin  — , 
4R8in— ,  4R8'n  -• 

.  a  .        8 

72.  If  A  is  a  right  angle,  i\  +  r^  =  c. 


164 


PLANE  TRIGOyoMKrii  ?  . 


M 


I 


I  =.■  ! 


73.  In  an  equilateral  triangle  .'i  H  =  (}r  =  2r,. 

74.  If  r,  r„  r,,  r,,  denote  the  radii  of  the  inscribed  and 
escribed  circles  of  a  triangle, 

tan»^  =  ^. 
2      r^rs 

75.  The  sides  of  a  triangle;  a.vv  in  arithmetic  progression, 
and  its  area  is  to  that  of  an  ec^ni lateral  triangle  of  the  same 
perimeter  as  3  is  to  />.  Find  the  ratio  of  the  sides  and  the 
value  of  the  largest  angle.  Ann.  As  7,  5,  3 ;  120**. 

76.  If  an  equil.iteral  triangle  bo  described  with  its 
angular  points  on  the  sidtns  of  a  given  right  isosceles 
triangle,  and  one  side  parallel  to  the  hypotenuse,  its  area 
will  be  2 a*' sin' 15°  sill  60",  where  a  is  a  sitlo  of  the  given 
triangle. 

77.  Tf  h  be  the  dilTerenco  between  tlu'  sides  containing 
the  right  angle  of  a  right  triangle,  a!id  S  its  area,  the  diam- 
eter of  the  circumscribing  circle  =  V^"'  +  48. 

78.  Three  circles  touch  one  another  externally  :  prove 
that  the  sqmire  of  the  anui  of  the  triangle  formed  by  jo'ii- 
ing  their  centres  is  equal  to  the  product  of  the  sum  and 
product  of  their  radii. 

79.  On  the  sides  of  any  triangle  equilateral  triangles  are 
described  externally,  and  their  centres  are  joined :  prove 
that  the  triangle  thus  formed  is  equilateral. 

SO.  Tf  ()„  Oj,  O,  are  the  centres  of  the  escribed  circles  of 
a  tiiangle,  then  the  area  of  the  triangle  0,().jO.,  =  area  ol 

a         .         b 


triangle  A  IK)    1  -j- 


{' 


+ 


a  -f  ft  —  c  J 


+  — 


b  ■\-  c  —  n      a  -\-  ('  —  b 

81.    If  the  centres  of  the  three  escril)ed  circles  of  a  tri- 
angle are  joined,  then  the  area  of  the  triangle  thus  formed 


a.  'k' 


is    -  -,  where  r  is  the  radius  of  the  inscribed  circle  of  the 


2r 


original  triangle. 


at 

th( 

an 


so  LI  T  ION   OF  TK I  ANGLES. 


165 


CHAPTER  VII. 

SOLUTION  OF  TRIANGLES. 

107.  Triangles.  —  Ju  every  triangle  there  are  six  elements, 
the  three  sides  and  the  three  angles.  When  any  three  ele- 
ments are  given,  one  at  least  of  the  three  being  a  side,  thp 
other  three  can  be  calculated.  The  process  of  determining 
the  unknown  elements  from  the  known  is  called  the  solution 
of  triangles. 

NoTl.  —  If  the  three  angles  only  of  a  triungle  are  given,  it  ii  itnpouible  to  deter- 
mine the  Hide*,  for  there  in  an  infinite  number  of  trianglea  that  are  equiangular  to 
one  another. 

Triangles  are  divided  in  Trigonometry  into  right  and 
oblique.  We  shall  commence  with  right  triangles,  and  shall 
suppose  C  the  right  angle. 


RIGHT  TRIANGLES. 

106.  There  are  Four  Cases  of  Right  Triangles. 

I.  Given  one  side  and  the  hypotenuse. 

II.  Oiven  an  acute  angle  and  the  hypotenuse. 

III.  Oiven  one  side  and  an  acute  angle. 

IV.  Given  the  two  sides. 

Let  AHC  be  a  triangle,  rig'.it-angled 
at  C,  and  let  a,  h,  and  c,  as  before,  he 
the  sides  opposite  the  angles  A,  B, 
and  C,  resjx'ctively. 

The  fornmlie  for  the  solution  of  right  triangles  are  (1), 
(2),  (.•{)  of  Art.  94. 


Ii 


■I. 

I' 


160 


PLANK   THIQONOMKTH  V. 


109.   Case  I.  —  Given  a  side  and  the  hypotenuse,  as  a  and 
c;  to  Jind  A,  B,  b. 


We  have 


sin  A  = 


a 


.'.  log  sin  A  =  log  a  —  log  c, 
from  which  A  is  deterniiued  ;  then  B  =  \H)°  —  A. 

Lastly,  b  =  c  cos  A. 

.*.  log  b  =  log  c  -f  log  cos  A. 

Thus  A,  B,  and  b  are  determined. 

Ex.  1.    Given  a  =  o3G,  c  =  Ml ;  find  A,  K,  b. 


Sohdion  by  Natural  Functions. 


We  have 


sin  A  =  -  = 


.rmm7. 


I 


c      941 
From  a  table  of  natural  sines  we  find  that 

A  =  34"  43' 22".    .-.  B  =  r,r>M 6' 38". 
Lastly,  6  =  c  cos  A  =  941  x  .821918    - 

=  773.425. 

Logarithmic  Sohdion. 


log  sin  A  =  log  a  —  log  c. 

loga  =  2.7291 648 
log  c  =  2.9735890 

log  sin  A*  =  9.7555752 

.        .-.  A  =  34°  43' 22". 

'       .-.  B  =  55M6'38". 


log  b  =  log  c  -f-  log  cos  A. 

log  c  =  2.9735890 
logcos  A  =  9.9148283 

log  ^  =  2.88841 79 

.-.  &  =  773.424. 


Our  two  methods  of  calculation  give  results  which  do  not 
quite  agree.  The  discrepancies  arise  from  the  defects  of 
the  tables.  • 


*  'fen  In  added  so  a»  to  get  tli    titbuhir  lti|<aritbmi  (Art.  76). 


liiuuT  ritJAyuLJSs. 


167 


The  process  of  solution  by  natural  sines,  cosines,  etc.,  can 
be  used  to  advantage  only  in  cases  in  which  the  measures 
of  the  sides  are  suiall  numbers. 

We  might  have  det«rmined  b  thus : 


or  thus : 


6  =  V(('  — a)(c4-a); 
b  =  a  tan  B. 


NoTR.  —  It  is  generally  better  to  c-oinpiitn  all  the  reqiilred  parts  from  the  given 
ones,  so  ihiu  Ifitn  crrur  is  made  in  (lelerminliiK  uiie  jmrt,  tliut  errur  will  nut  ufTect  llii* 
coiiipiitution  of  tlio  other  parts. 

'I'u  lest  the  uceiiraoy  of  the  work,  vompule  the  same  parts  by  different  furmulw. 

Ex.  2.   (liven  a  =  21,  c=29;  iind  A,  H,  b. 

Ans.  A  =  4G°2.r50",  B  =  4.r  30' 10",  &  =  20. 

Note.  —  In  these  examples  the  student  must  tiud  the  necessary  logarithms  from 
the  tables. 


r 


110.   Case  II.  —  Qicen  an  acute  angle  and  the  hypotenuse, 
UH  A  and  c ;  tojiiul  B,  a,  b. 

We  have  B  =  90°  -  A.  ^ 

Also  a  =  csinA,  and  6  =  ccosA. 

.•.  log  a  =  log  c  -f-  log  sin  A  ; 
and  log  b  =  log  c  -\-  log  t!OS  A. 

Thus  H,  a,  and  b  are  determined. 

Ex.  1.    Given       A  =  54°  28',  c  =  125 ;  find  B,  a,  b. 
B  =  90°  -  A  =  35°  32'. 


Solution  by  Natural  Functions. 

We  have  o  =s  c  sin  A,  and  b  =  c cos  A. 

Using  a  table  of  natural  sines,  we  have 

a  =  125  X  .813778  =  101.722, 
and  6  =  126  X  .581177  =    72.(517. 


'•    / 


168 


PLANK  TUWONOMKTHY. 


Loyurithmic  Solution. 


log  a  =  log  c  4-  log  sill  A. 

logc  =  2.0909100 
log  sin  A  =  9.9105057 

loga*  =  2.0074157 

.-.  0  =  101.722. 


log  b  =  log  c  +  log  cos  A. 

log  0  =  2.09(59100 
log  008  A  =  9.7(>4.'i0.S0 

iog6*  =  l.S()12180 

.-.  6  =  72.047. 


Ex.  2.   Given  A  =  37°  10',  c  =  8702 ;  find  a  and  6. 

Ans.  5293.4;  0982.3. 

HI.  Case  in.  —  Oiveyi  a  aide  and  an  acute  angle,  as  A 
and  a;  tojind  B,  b,  c. 


We  have 

K  =  90*  -  A. 

Also 

h-      ^        and  r-     " 

y  —  ,■    -  . ,  aiiu  c  —    . 
tan  A                  sin  A 

.'.  log  b  =  log  a  —  log  tan  A, 
and  log  c  =  log  a  —  log  sin  A. 

Ex.  1.   Given  A  =  32°  15'  24",  a  =  5472.5 ;  find  B,  b,  c. 

Solution.  • 

B  =  90°  -  A  =  57°  44' 30". 


log  b  =  log  a  —  log  tan  A. 

log  a  =  3.7381858 
log  tan  A  =  9.8001090 

log  6  =  3.9380768 

.-.  &  =  8071.162. 


log  c  =  log  a  —  log  sin  A. 

log  a  ■-■=  3.7381858 
log  sin  A  =  9.7273070 

log  (J  =  4.0108782 

.-.  c  =  10253.04. 


Ex.  2.    Given  A  =  34°  18',  a  «  337.0  j  find  B,  b,  c.      " 

Ana.  B  =  55°42';  6  =  348.31;  c  =  421.03. 


*  Ten  !•  r^tcted  becauite  the  Ubuiat  .ugarUtiiuic  (uucliuiu  are  tou  large  by  tea 
(Art  76). 


HIGHT  TllIANGLES. 


1U9 


112.  Case  IV. 
A,  B,  c. 

We  have 
Also 


Given  the  two  sides,  as  a  and  b;  to  find 


tan  A 


a 


then  B  =  90°  -  A. 


c  =  a  oosec  A  = 


a 


sin  A 
.'.  log  tan  A  =  lo{^  a  —  loj,'  h, 

and  logc  =  loga  —  lo}j;sin  A. 

Ex.   Given  a  =  22GG.35,  h  =  r»4;«).24 ;  find  A,  B,  c. 

Solution. 


log  tan  A  =  log  a  —  log  b. 

log  a  =  ;3.35rK{270 
log  b  =  ;i7355382 

log  tan  A  =  9.6197888 

.-.  A  =  22"  37' 12". 

.-.  B  =  07°  22' 48". 


log  c  =  log  a  —  log  sin  A. 

log  a  =  3.3553270 
logsinA  =  9.r»8r)02(;() 

log  c  =  3.7703004 

.-.  0  =  5892.51. 


NoTK.  —  In  thii  exnmple  we  miKht  have  found  r  by  nioana  of  the  formula 
C-Va'  +  fc';  but  wc  would  liuvo  had  to  go  through  thi'  jtrocess  of  aquariny  the 
viihiea  of  a  and  b.  If  these  vulueM  are  Dimple  nuiiibiTH,  it  U  often  euMJcr  to  lliid  c  In 
this  way;  but  this  value  of  c  is  not  adapted  to  loKarithnm.  A  fonnula  which  uonsists 
entirely  oT  fnctora  is  always  preferred  to  one  wkich  consists  uf  termn,  when  any  of 
those  terms  contain  any  power  of  the  quantities  involved. 

113.  When  a  Side  and  the  Hypotenuse  are  nearly  Equal. 
—  When  a  side  and  the  hypotenuse  are  given,  as  a  and  c 
in  (!ase  I.,  and  are  nearly  equal  in  value,  the  angle  A  is  very 
near  90°,  and  cannot  be  determined  with  much  accuracy 
from  the  tables,  because  the  sines  of  angles  near  90°  differ 
very  little  frr»m  one  another  (Art.  81).  It  is  therefore 
desirable,  in  this  case,  to  find  B  first,  by  either  of  the 
following  formultB . 


sm 


.    B        /l— cosB  /.    ,    wn\ 

^^2=\ 2 — (Art.  oO) 

(1) 


■V 


c  —  a 
2c 


170  PLANE  TIlWOyOMETUr. 

tan2=Jp^^ (Art.  CO) 

=J"p (2) 

Then  6  =  cco8A (3) 

or  =V{,7+a){c-a) (4) 

Ex.  1.    Given  a  =  4l>0".2105y,  c  =  4(>02.836;  find  B. 
c  -  a  =  ().Cli541,  log  (c  -  a)  =  i.7yr.lG48 

2  e  =  9206.072,  log  2  c  =  3. 1)040555 

2)5.8321093 
log  sin  1^  =  7.9100547 

B  =  50'40".30.  .'.   ~  =  2S'20".18. 

NoTK.  — Thu  characterlatlc  5  In  increaited  numerically  to  6  to  muke  ll  dlvlHible 
by  'I  (see  Note  4  of  Alt.  00).  Teu  in  then  added  to  the  churact«ri«tiu  3,  luuklntc  it  7, 
«o  ua  to  agree  with  tbe  Table*  (Art.  70). 

There  is  a  slight  error  in  the  above  value  of  B  on  account 
of  the  irregular  differences  of  the  log  sines  for  angles  near 
0°  (Art.  81).  A  more  accurate  value  may  he  found  by  the 
principle  that  the  sines  of  small  angles  are  approximately 
proportional  to  the  angles  (Art.  130). 

EXAMPLES. 

The  following  right  triangles  must  be  solved  by  log- 
arithms. 

1.  Given  «  =  (M),  c  =  100 ;  find  A,  B,  h. 

An».  A  =  30°  ",2' ;  B  =  53°  8'  j  6  =  80. 

2.  Given  a=i;57.0<),  (-  =  240;  fiiid  A,  B,  &. 

An».  A  =  :;.r ;  B  =  55°  ;  /*  =  100.59. 


niOUT  TRIANOLES. 


171 


3.  Given  a=  147,  c=  KS4;  Hiul  A,  IJ,  b. 

Ans.  A=:5n'ao";  IJ  =  oG°  58' 1'5" ;  6  =  llO.G7. 

4.  Given  a  =  10(),  c  =  2(M) ;  tind  A,  li,  6. 

AiiB.  A  =  30°;  H  =  ()()°;  h=U)OV^. 

6.   Given  A  =  40'*,  c=  100;  find  1),  a,  b. 

Anit.  li  =  50'';  (1  =  04.279;  6  =  70.004. 

6.  Given  A  =  30°,  c  =  160 ;  find  B,  o,  b. 

Ans.  H  =  60°;  a  =  7r»;  6  =  75V3. 

7.  Given  A  =  32°,  (•=  1700;  find  IJ,  a,  6.  '•  ■■    •     ' 

ylHS.  H  =  r>«°;  a  =  l>32.00;  />=  1492.57. 

8.  Given  A  =  35°  10'  25",  e  =  072.3412 ;  find  B,  «,  6. 

ytn«.   H  =  54°  43' 35";  a  =  388.20;  6  =  548.9. 

9.  Given  A  =  75°,  a  =  80;  find  B,  6,  c. 

ylus.   H  =  15°;  6  =  80(2-V3);  c  =  80(  VO -V2). 

10.  Given  A  =  30°,  a  =  520 ;  find  li,  6,  c. 

yln«.  B  =  54°;  6  =  716.72;  0  =  884.08. 

11.  Given  A  =  34°  15',  a  =  843.2;  find  B,  6,  c. 

^ns.  B  =  55°  45' ;  c  =  1498.2. 

12.  Given  A  =  07°  37'  15",  b  =  254.73 ;  find  B,  a,  e. 

Ans.  B  =  22°  22'  46" ;  a  =  018.00 ;  c  =  009.05. 

13.  Given  a  =  75,  6  =  75;  find  A,  B,  c. 

Ans.  A  =  45°  =  B  J  c  =  75 V2. 

14.  Given  a  =  21,  6  =  20 ;  find  A,  B,  c. 

Ans.  A  =  40°  23' 60";  c  =  29. 

15.  Given  a  =  .100.43,  6  =  500 ;  find  A,  B,  c. 

J  Ans.  A  =  31° ;  B  =  59° ;  c  =  683.31. 

16.  Given  a  =  4846,  6  =  4742;  find  A,  B,  c. 

Ans.  A  =  45°  30'  50". 


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172 


PLANE  TRIGONOMETRY. 


■iH--  !'v"l 


OBLIQUE  TRIANGLES. 

114.  There  are  Four  Cases  of  Oblique  Triangles. 

I.  Given  a  side  and  two  angles. 

II.  Given  two  sides  and  the  angle  opposite  one  of  them. 

III.  Given  two  sides  and  the  included  angle. 

IV.  Given  the  three  sides. 

The  formulae  for  the  solution  of  oblique  triangles  will  be 
taken  from  Chap.  VI.  Special  attention  must  be  given  to 
the  following  three,  proved  in  Arts.  95,  96,  97. 

a  b  c 


(1)  The  Sine-rule,     ; 

(2)  The  Cosine-rule, 

(3)  The  Tangent-rule, 


sin  A 
cos  A: 


sin  B      sin  C 
62  +  c-2  -  g^ 
26c 


A-B      a-b     .C 
tan = cot  — 

2  a  +  b       2 


115.   Case  I.    Given  a  side  and  two  angles,  as  a,  B,  C;  Jind 
A,  b,  c. 


(1) 

A  =  180° - 

(2) 

b            a 

sin  B      sin  A 

r.s^ 

c            a 

(B  +  C). 


siuC      sin  A 
These  determine  b  and  c. 


A  is  found. 

,      a  sin  B 
0  = 

sm  A 

asinC 
c  =  — ; 

sin  A 


Ex.  1.   Given  a  =7012.6,  B  =  38°  12' 48",  C=60°;  find 
A,  b,  c. 

Solution. 

A  =  180°  -  (B -f- C)  =  81°  47' 12". 


log  a  =  3.8458729 

log  sin  B  =  9.9714038 

colog*sinA  =  0.0044775 

log  b  =  3.82r7542 

.-.  6  =  6633.67. 


log  a  =  3.8458729 

log  sin  C  =  9.9375306 

colog  sin  A  =  0.0044775 

log  c  =  3.7878810 

.-.  c  =  6135.94. 


*  See  Art.  00. 


J31 


OBLIQUE  TRIANGLES. 


173 


Ex.  2.    Given  a=1000,  1^=45°,  C  =  127°  19' ;  find  A,  b,  c. 
Ans.  A  =  7°  41' ;  b  =  5288.8 ;  c  =  5948.5. 

116.  Case  II.    Given  two  sides  and  the  angle  opposite  one 
of  them,  as  a,  b,  A;  find  B,  C,  c. 

(1)  sin  B  =  — ;  thus  B  is  found. 


a 


(2) 
(3) 


C  =  180°  -  (A  +  B);  thus  C  is  found, 
a  sin  C 


c  = 


sin  A 


;  thus  c  is  found. 


This  is  usually  known  as  the  ambiguous  case,  as  shown  in 
geometry  (B.  II.,  Prop.  31).  The  ambiguity  is  found  in 
the  equation 

b  sin  A 


sinB 


a 


Since  the  angle  is  determined  by  its  sine,  it  admits  of 
two  values,  which  are  supplements  of  each  other  (Art.  29). 
Therefore,  either  value  of  B  may  be  taken,  unless  excluded 
by  the  conditions  of  the  problem. 

I.  If  a  <  6  sin  A,  sin  B  >  1,  which  is  impossible ;  and 
therefore  there  is  no  triangle  with  the  given  parts. 

II.  If  a  =  6  sin  A,  sin  B  =  1,  and  B  =  90°;  therefore  there 
is  one  triangle  —  a  right  triangle  —  with  the  given  parts. 

III.  If  a  >  6  sin  A,  and  <  6,  sin  B  <  1 ;  hence  there  are 
two  values  of  B,  one  being  the  supplement  of  the  other, 
i.e.,  one  acute,  the  other  obtuse,  and  both  are  admissible ; 
therefore  there  are  two  triangles  with  the  given  parts. 

IV^.  If  a>b,  then  A  >  1^,  and  since  A  is  given,  B  must 
be  acute;  thus  there  is  only  one  triangle  with  the  given 
parts.  ■  •.. 


174 


PLANE  TRIGONOMETRY. 


Tliese  four  cases  may  be  illustrated  geometrically. 

Draw  A,  the  given  angle.     Make  AC  =  b ;  draw  the  per- 
pendicular CD,  which  =  b  sin 
A.    With  centre  C  and  radius 
a,  describe  a  circle. 

I.  If  a<b  sin  A,  the  circle 
will  not  meet  AX,  and  there- 
fore no  triangle  can  be  formed 
with  the  given  parts. 

II.  If  a  =  6  sin  A,  the  cir- 
cle touches  AX  in  B' ;  there- 
fore   there  is   one  triangle, 
right-angled  at  B. 

III.  If  a>b  sin  A,  and 
<  6,  the  circle  cuts  AX  in 
two  points  B  and  B',  on  the  \y^ 


same  side  of  A ;  thus  there  ^  \    A  D  /'g 

are  two  triangles  ABC  and  AB'C,  each  having  the  given 
parts,  the  angles  ABC,  AB'C  being  supplementary. 

IV.  If  a>b,  the  circle  cuts  AX  on  opposite  sides  of  A, 
and  only  the  triangle  ABC  has  the  given  parts,  because  the 
angle  B'AC  of  the  triangle  AB'C  is  not  the  given  angle  A, 
but  its  supplement. 

These  results  may  be  stated  as  follows :        , 

a<&sinA,  no  solution. 

ax=  &sin  A,  one  solution  (right  triangle). 

a  >  6  sin  A  and  <  6,  two  solutions. 

a^b,  one  solution. 

These  results  may  be  obtained  algebraically  thus : 

We  have         a''  =  ?>-  +  c-  -  2  6c  cos  A     .     .     .     (Art.  Ofi) 


.-.  c=b  cos  A  ±  Va^  —  b^  sin^  A, 


OBLIQUE  TBI  ANGLES. 


175 


giving   two   roots,   real  and   unequal,  equal  or  imaginary, 
according  as  a  >,  =,  or  <  6  sin  A. 

A  discussion  of  these  two  values  of  c  gives  the  same 
results  as  are  found  in  the  above  four  cases.  We  leave 
the  discussion  as  an  exercise  for  the  student. 

Note.  —  When  two  sides  and  the  angle  opposite  the  greater  are  given,  there  can 
be  no  ambiguity,  for  the  angle  opposite  the  less  must  be  acute. 

When  the  given  angle  is  a  right  angle  or  obtuse,  the  other  two  angles  arc  both 
acute,  and  there  can  be  no  ambiguity. 

In  the  solution  of  trianglen  there  can  be  no  ambiguity,  except  when  an  angle  is 
determined  by  the  sine  or  cosecant,  and  in  no  case  whatever  when  the  triaugle  has 
a  right  angle. 

Ex.  1.   Given  a  =  7,  &  =  8,  A  =  27°  47' 45";  find  B,  C,  c. 


Solution. 


log  6  =  0.9030900 

log  sin  A  =  9.66868G0 

colog  a  =  9.1549020 


log  a  =  0.8450980 

log  sin  C  =  9.9375306 

colog  sin  A  =  0.3313140 

log  c=  1.1139426 


log  sin  B  =  9.7266780 

.-.  B  =  32°12'15",orl47°47'45".  .:  c  =  13. 

.-.  C  =  120°,  or  4"  24' 30". 

Taking  the  second  value  of  C  as  follows : 

...    .,  log  a  =  0.8450980 

log  sin  C  =  8.8857232        ' 
■  colog  sin  A  =  0.3313140  '  > 

log  c  =  0.0621352  '       .      ■ 

"        i  .-.  c  =  1.1538. 

Thus,  there  are  two  solutions.     See  Case  III. 

Ex.2.   Given  a  =  31.239,  &  =  49.5053,  A  =  32°  18';   find 
B,  C,  c.  Ans.  B  =  56°  56'  56".3,  or  123°    3'    3".7 ; 

C  =  90°45'    3". 7,  or    24°38'56".3; 
c=  58.456,  or  24.382. 


n 


.«> 


7'  ^  " !  » 


176 


PLANE   TRIG ONOMETR  Y. 


117.   Case  III.  —  Given  two  sides  and  the  included  angle, 
as  a,  6,  C ;  find  A,  B,  c. 


(1) 


tan 


A-B      a 


a-\-b 


b      .C 
-  cot  — 

2 


(Art.  114) 


Hence  — - —  is  known,  and  — 4^—  =  90'' 


.'.  A  and  B  are  found 
(2) 


a  sinC 
c  =  — - — — t  or 


2 

b  sinC 


C 
2* 


sin  A  '         sin  B  ' 

and  thus  c  is  found  and  the  triangle  solved. 

In  simple  cases  the  third  side  c  may  be  found  directly  by 
the  formula 

c=:-\/aF+¥'-2ab'GosC   .     .     .     (Art.  96) 

or  the  formula  may  be  adapted  to  logarithmic  calculation 
by  the  use  of  c  subsidiary  angle  (Art.  90). 

Ex.  1.    Given   a  =  234.7,   b  =  185.4,  C  =  84°  36' ;   find  A, 

.       ,  Solution. 

log{a  -  h)  =    1.6928460 


a  =  234.7 
b  =  185.4 


a-b=   49.3 
a  +  b  =  420.1 

.-.  ^  =  42°  18'. 

^^4^ -=47'' 42'. 
2 

.-.  A  =  55°    2' 56", 
B  =  40°21'   4", 
C  =  84°  36', 
c  =  286.0746. 


coiog(a  +b)=    7.3766473 

log  cot  ^  =  10.0409920 

^        2      


log  tan 


B 


2 
A-B 


=   9.1104862 
=  7°  20'  56". 


log  b  =  2.2681097 

log  sin  0  =  9.9980683 

colog  sin  1?  =  0.1887804 

log  c  =  2.4549584 


OBLIQUE  TRIANGLES. 


177 


Ex.  2.  Given  a  =■-  .062387,  b  =  .023475,  C  =  110°  32' ;  find 
A,  B,  c. 

Ans.  A  =  52°  10'  33" ;  B  =  17°  17'  27" ;  c  =  .0739635. 

118.  Case  IV.   Given  the  three  sides,  as  a,  b,  c;Jind  A,  B,  C. 

The  solution  in  this  case  may  be  performed  by  the  for- 
mulae of  Art.  99.  By  means  of  these  formulae  we  may 
compute  two  of  the  angles,  and  find  the  third  by  subtract- 
ing their  sura  from  180°.  But  in  practice  it  is  better  to 
compute  the  three  angles  independently,  9,nd  check  the 
accuracy  of  the  work  by  taking  their  sum. 

If  only  one  angle  is  to  be  found,  the  formulae  for  the  sines 
or  cosines  may  be  used.  If  all  the  angles  are  to  be  found, 
the  tangent  formulae  are  the  most  convenient,  because  then 
we  require  only  the  logarithms  of  the  same  four  quantities, 
s,  s  —  a,  s  —  b,  s  —  c,to  find  all  the  angles ;  whereas  the  sine 
and  cosine  formulae  require  in  addition  the  logs  of  a,  b,  c. 

The  tangent  formulae  (  A.rt.  99)  may  be  reduced  as  follows : 


tan- 


\       s(s 


-b)(s-c) 
{s  —  a) 


'■t  i;  ' 


,      A        r 
tan— =  — 
o 


^     1        j{s-a)(s~b)(s-c)       vK- 

5  —  Ct    «  S 

(Art.  102) 


s  —  a 


14 


B         T 

Similarly,         tan—  = -? 

2      s  —  b 

V    '    '        tan  — = 


2     s-c 


Note.  —  The  quantity  r  is  the  radius  of  the  inscribed  circle  (Art.  102). 


V  '■  V, 


•^w — T=- 


178 


PLANE  TRIGONOMETRY. 


■  W. 


-.1 


I: 


Ex.  1.   Given  a  =  13,  6  =  14,  c  =  15 ;  find  A,  B,  C. 


Solution. 


I    -'  I 


a  =13 
6  =  14 
c  =  15 

2s  =  42 

5=21. 

s  -a  =  8, 
5-6  =  7,* 
5—  c=6. 


log(s-a)=    .9030900 

log(s-6)=    .8450980 

log(s-c)=    .7781513 

colog  5  =  8.6777807 

log  7-2  =  1.2041200 

logr=    .6020600. 

.-.  log  tan -  =  9.6989700. 
^        2 

.-.  log  tan -  =  9.7569620; 

.-.  log  tan  ^  =  9.8239087.  ■ 

.-.  A  =  53°7'48".38; 

B  =  59°  29' 23".18 ;  /. 

C  =  67°  22' 48".44.  v  -'uv,.,, 

Without  the  use  of  logarithms,  the  angles  may  be  found 
by  the  cosine  formulae  (Art.  96).  These  -.nay  sometimes 
be  used  with  advantage,  when  the  given  lengths  of  a,  b,  c 
each  contain  less  than  three  digits. 

Ex.  2.  Find  the  greatest  angle  in  the  triangle  whose 
sides  are  13,  14,  15. 

Let  a  =  15,  6  =  14,  c  =  13.     Then  A  is  the  greatest  angle. 

Then       oosX=='l±^  =  ^,±§^f 

26c  2x14x13 

=  ^  =  .384615  =  cos  67°  23',  nearly 
(by  the  table  of  natural  sines). 

.'.  the  greatest  angle  is  67°  23'. 


EXAMPLES. 


179 


EXAMPLES. 

1.   Given  a  =  254,  B  =  16°,  C  =  64°;  find  6  =  71.0919. 
=  338.65,    A  =  53°  24',   B  =  66°  27'j    find 


2.  Given    c 
«  =  313.46. 

3.  Given    c 
b  =  31.43. 

4.  Given    a 
b  and  c. 

5.  Given  a  = 


6.    Given  a  = 


7.   Given  a  = 


8.    Given  a  = 


9.  Given  6  = 

10.  Given  6  = 

11.  Given  a  = 

12.  Given  a  = 


=  38,    A  =  48°,    B  =  64°;    find    a  =  28.87, 

=  7012.5,    B  =  38°  12' 48",    C  =  60°;    find 
Aits.  6  =  4382.82;  c  =  6135.94. 

528,  6  =  252,  A  =  124°  34';  find  B  and  C. 
Ans.  B  =  23°8'33";  C  =  32°  17' 27". 

170.6,  6  =  140.5,  B  =  40°;  find  A  and  C. 
Ans.  A  =  51°  18'  21",  or  128°  41'  39" ; 
C  =  88°  41' 39",  or    11°  18' 21". 

97,  6  .=  119,  A  =  50° ;  find  B  and  C. 

Ans.  B  =  70°   0'  56",  or  109°  59'   4" ; 
C  =  59°59'   4",  or    20°   0'56". 

7,  6  =  8,  A  =  27° 47' 45";  find  B,  C,  c. 
A71S.  B  =    32°  12'  15",  or  147°  47'  45" ; 
C  =  120°,  or      4°  24' 30"; 

c  =  13,  or  1.15385.  ,      , 

55,  c  =  45,  A  =  6° ;  find  B  and  C. 

Ans.  B=  149°  20' 31";  C  =  24°  39' 29". 

131,  c  =  72,  A  =  40° ;  find  B  and  C. 
Ans.  B  =  108°  36' 30"  ;  C  =  31°  23' 30". 

35,  6  =  21,  C  =  50°;  find  A  and  B. 

Ans.  A  =  93°  11' 49";  B  =  36°  48' 11". 

601,  6=289,  C=100°  19'  6";  find  A  and  B. 
Ans.  A  =  56°  8' 42"  5  B  =  23°  32' 12", 


ill 


Ui' 


mi 


iiiil' 


180  ,  PLANE  TRIGONOMETRY. 

13.  Given  a  =  222,  h  =  318,  c  =  406 ;  find  A  =  32°  57'  8". 

14.  Given  a  =  275.36,  &=  189.28,  c=301.47  ;  find  A,  B,  C. 
Ans.  A  =  63°  30' 57";  B  =  37°  58' 20";  C  =  78°  30' 43". 

16.   Given  a  =  5238,  6  =  5662,  c  =  9384 ;  find  A  and  B. 

Ans.  A  =  29°  17' 16";  B  =  31°  55' 31". 

16.    Given  a  =  317,  h  =  533,  c  =  510 ;  find  A,  B,  C. 

Ans.  A  =  35°  18'  0" ;  B  =  76°  18'  52" ;  C  =  68°  23'  8". 

119.  Area  of  a  Triangle  (Art.  101). 


rind  the  area : 

1.  Given  a  = 

2.  Given  a  = 

3.  Given  &  = 

4.  Given  a  = 
6.  Given  6  = 

6.  Given  a  = 

7.  Given  a  = 

8.  Given  a  = 

9.  Given  a  = 

10.  Given  a  = 

11.  Given  a  = 

12.  Given  a  = 


EXAMPLES. 

116.082,  6  =  100,  C  =  118°  15' 41". 

Ans.  5112.25. 
8,  &  =  5,  C  =  60°.  17.3205. 

21.5,  c  =  30.456,  A  =  41°  22'.         216.372. 

72.3,  A  =  52°  35',  B  =  63°  17'.       2644.94. 

100,  A  =  76°  38'  13",  C  =  40°  5'.   3506.815. 

31.325,  B  =  13°57'2",  A  =  53°  11' 18". 

Ans.  135.3545. 

.582,  6  =  .601,  c  =  .427.  .117655. 

408,  6  =  41,  c  =  401.        ■  '         8160. 

.9,  6  =  1.2,  c  =  1.5.  .64. 

21,  6  =  20,  c  =  29.  210. 

24,  6  =  30,  c  =  18.        -^^  -         216. 

63.89,  6  =  138.24,  c  =  121.15.  3869.2. 


HEIGHT  OF  AN  OBJECT. 


181 


'i  ; 


MEASUREMENT   OF    HEIGHTS   AND   DISTANCES. 

120.  Definitions.  —  One  of  the  most  important  applica- 
tions of  Trigonometry  is  the  determination  of  the  heights 
and  distances  of  objects  which  cannot  be  actually  measured. 

The  actual  measurement,  with  scientific  accuracy,  of  a 
line  of  any  considerable  length,  is  a  very  long  and  difficult 
operation.  But  the  accurate  measurement  of  an  angle,  with 
proper  instruments,  can  be  made  with  comparative  ease  and 
rapidity. 

By  the  aid  of  the  Solution  of  Triangles  we  can  determine : 

(1)  The  distance  between  points  which  are  inaccessible. 

(2)  The  magnitude  of  angles  which  cannot  be  practically 
observed.  .  -  ; 

(3)  The  relative  heights  of  distant  and  inaccessible 
points. 

A  vertical  line  is  the  line  assumed  by  a  plummet  when 
freely  suspended  by  a  cord,  and  allowed  to  come  to  rest. 

A  vertical  plane  is  any  plane  containing  a  vertical  line. 

A  horizontal  plane  is  a  plane  perpendicular  to  a  vertical 
line. 

A  vertical  angle  is  one  lying  in  a  vertical  plane.  ; 

A  horizontal  angle  is  one  lying  in  a  horizontal  plane. 

An  angle  of  elevation  is  a  vertical  angle  having  one  side 
horizontal  and  the  other  ascending. 

An  angle  of  depression  is  a  vertical  angle  having  one  side 
horizontal  and  the  other  descending.  '      '  "  .  , 

By  distance  is  meant  the  horizontal  distance,  unless  other- 
wise named. 

By  height  is  meant  the  vertical  height  above  or  below  the 
horizontal  plane  of  the  observer. 

For  a  description  of  the  requisite  instruments,  and  the 
method  of  using  them,  the  student  is  referred  to  books  on 
practical  surveying.*  • 

*  See  Johnson's  Surveying,  Gillespie's  Surveying,  Clarke's  Geodesy,  Gore's 
Geodesy,  etc. 


ill 


Hi 


mi 


182 


PLANE  TBIOONOMETRY, 


^y\ 


^ 


121.  To  find   the   Height  of   an  Object   standing  on  a 
Hcrizontal  Plane,  the  Base  of  the  Object 
being  Accessible. 

Let  BC  be  a  vertical  object,  such  as 
a  church  spire  or  a  tower. 

From  the  base  C  measure  a  horizon-  /^ 
tal  line  CA. 

At  the  point  A  measure  the  angle  of  elevation  CAB, 

We  can  then  determine  the  height  of  the  object  BC ;  for 

BC  =  AC  tan  CAB. 


.^ 


EXAMPLES. 

1.  If  AC  =  100  feet  and  CAB  =  60°,  find  BC.  '  •' 

Ans.  173.2  feet. 

2.  If  AC  =  125  feet  and  CAB  =  52°  34',  find  BC. 

Ans.  163.3  feet. 

3.  AC,  the  breadth  of  a  river,  is  100  feet.  At  tlie  point 
A,  on  one  bank,  the  angle  of  elevation  of  B,  the  top  of  a  tree 
on  the  other  bank  directly  opposite,  is  25°  37';  find  the 
height  of  the  tree.  Ans.  47.9  feet. 

122.  To  find  the  Height  and  Distance  of  an  Inaccessible 
Object  on  a  Horizontal  Plane. 

Let  CD  be  the  object,  whose  base  D 

is  inaccessible ;  and  let  it  be  required  /^  / 

to  find  the  height  CD,  and  its  horizon-  x^    /\ 

tal  distance  from  A,  the  nearest  acces-  B     ^  '  ^ — ^ 
sible  point. 

(1)  At  A  in  the  horizontal  line  BAD  observe  the  Z  DAC 
=  a ;  measure  AB  =  a,  and  at  B  observe  the  Z  DBC  =  ^. 


/ 


Then 


CA  =  -^"^"^ 
sin  (a  —  /8) 


(Art.  95) 


/ 


HEIGHT  OF  AN  OBJECT. 


183 


a 


■i 


B 


and' 


rn^      r^  K    •           asinrtsinjS 
,•.  CD  =  CAsin  u=        ^, 

8in(a  — /?) 

Kr\       An«„„         a  sin  yS  cos  a 
AD  =  AC  cos  a  = ^ • 

sin(rt  — /3) 


rr.i.T 


■):,Yi 


(2)     When  the  line  BA  cannot  be  yneasured  directly  toward 
the  object. 

At  A  'Observe  the  vertical  Z  CAD 
=  a,  and  the  horizontal  Z  DAB  =  fi ; 
measure   . AR  =  a,  and  at   B   observe    y^ 
the  /L  DBiV  =  y. 

Then 


/ 


f. 

\''/^ 

AD-    "''^'^r   . 

0^ 

sin(y3  +  y) 

CD  =  AD  tan  a 

_  a  sin  y  tan  a 

EXAMPLES. 

1.  A  river  300  feet  wide  runs  at  the  foot  of  a  tower, 
which  subtends  an  angle  of  22°  30'  at  the  edge  of  the  remote 
ban V ;  find  the  height  of  the  tower.  Ans.  124.26  feet. 

2.  At  3G0  feet  from  the  foot  of  a  steeple  the  elevation  is 
half  what  it  is  at  135  feet ;  find  its  height.      Ans.  180  feet. 

3.  A  person  standing  on  the  bank  of  a  river  observes  the 
angle  subtended  by  a  tree  on  the  opposite  bank  to  be  60°, 
and  when  he  retires  40  feet  from  the  river's  bank  he  finds 
the  angle  to  be  30° ;  find  the  height  of  the  tree  and  the 
breadth  of  the  river.      ,  ,    ^ws.  20V3;  20. 

4.  What  is  the  height  of  a  hill  whose  angle  of  elevation, 
taken  at  the  bottom,  was  46°,  and  100  yards  farther  off,  on 
a  level  with  the  bottom,  the  angle  was  31°  ? 

;      Ans.  143.14  yards. 


184 


PLANE  TRIGONOMETRY. 


123.  To  find  the  Height  of  an  Inaccessible  Object  sitiiated 
above  a  Horizontal  Plane,  and  its  tj 

Height  above  the  Plane. 

Let  CD  be  the  object,  and  let 
A  and  B  be  two  points  in  the 
horizontal  plane,  and  in  the  same 
vertical  plane  with  CD. 

At   A,  in  the    horizontal   line     B~ 
BAE,    observe    the    A  C AE  =  «,  , 

and   DAE  =  y ;    measure    AB  =  a,   and  at   B   observe  the 
ZCBE  =  /8.  ' 


'  <^ 

c 

,   ■  . 

yy 

.     .  ■■  .. 

D 

• 

y^' 

<<'<' 

,./-?  . 

\ 

\ 

3         a     A 

E 

Then 

Also, 


^^^^a^asuij (Art.  122) 

sin(a  — /8) 

^^^ocos^sin^ (Art.  122) 

sin  {a—  ft) 

r)F  _  ^  ^os  a  sin  ft  tan  y 

~       sin  (a  -"ft)      '  \ 

CD  = ^-    )sin  a  —  cos  a  tan  -y  ( 

sm(a-ft)^  ^^ 

_  g sin/? sin  {a  —  y)        ' 

cos  y  sin  {a  —  ft)  ''•'^ 


EXAMPLES. 

1.  A  man  6  feet  high  stands  at  a  distance  of  4  feet  9 
inches  from  a  lamp-post,  and  it  is  observed  that  his  shadow 
is  19  feet  long :  find  the  height  of  the  lamp.      Ans.  7^  feet. 

2.  A  flagstaff,  25  feet  high,  stands  on  the  top  of  a  cliff, 
and  from  a  point  on  the  seashore  the  angles  of  elevation 
of  the  highest  and  lowest  points  of  the  flagstaff  are  observed 
to  be  47°  12'  and  45°  13'  respective!  ■  find  the  height  of  the 
cliff.  .   .   ,  '        .  ,  Ans.  348  feet. 

3.  A  castle  standing  on  the  top  of  a  cliff  is  observed 
from  two  stations  at  sea,  which  are  in  line  with  it ;  their 


HEIGHT  OF  AN  OBJECT. 


185 


J. 


distance  is  a  quarter  of  a  mile :  the  elevation  of  the  top 
of  the  castle,  seen  from  the  remote  station,  is  16°  28';  the 
elevations  of  the  top  and  bottom,  seen  from  the  near 
station,  are  52°  24'  and  48°  38'  respectively:  (1)  what  is 
its  height,  and  (2)  what  its  elevation  above  the  sea? 

^ns.   (1)  60.82  feet;   (2)  445.23  feet. 

124.  To  find  the  Distance  of  an  Object  on  a  Horizontal 
Plane,  from  Observations  made  at  Two  Points  in  the  Same 
Vertical  Line,  above  the  Plane. 

Let  the  points  of  observation  A 
and  B  be  in  the  same  vertical  line, 
and  at  a  given  distance  from  each 
other ;  let  C  be  the  point  observed, 
whose  horizontal  distance  CD  and   -s^l 

C  D 

vertical  distance  AI)  are  required. 

Measure  the  angles  of  depression,  6BC,  aAC,  equal  to  a 
and  /8  respectively,  and  denote  AB  by  a. 


Or— 


B 
a 
A 


Then 


and 


BD  =  CDtan«,  AD  =  CD  tan /3. 

.'.  a  =  CD  (tan  «  —  tan /3) . 
^^^^  _  ^  cos  a  cos  ji 

sin  («  —  /3)  ' 

a  cos  a  sin  ^ 

sin  (rt  —  (i) 


A  jv u,  i;us  «  sill  p 


ill' 


I  !in  I 


i^^ii 


^lili 


'ill' 

lii! 


EXAMPLES. 

1.  From  the  top  of  a  house,  and  from  a  window  30  feet 
Ijelow  the  top,  the  angles  of  depression  of  an  object  on  the 
ground  are  15°  40'  and  10°  :  find  (1)  the  horii,ontal  distance 
of  the  object,  and  (2)  the  height  of  tlie  house. 

Ans.   (1)  288.1  feet ;  (2)  80.8  feet. 

■  2.  From  the  top  and  bottom  of  a  castle,  which  is  68  feet 
high,  the  depressions  of  a  ship  at  sea  are  observed  to  be 
16°  28'  and  14° :  find  its  distance.  \ns.  570.2  yards. 


186 


PLANE  TRIGONOMETHY. 


125.  To  find  the  Distance  between  Two  Inaccessible  Objects 
on  a  Horizontal  Plane. 

Let  C  and  D  be  the  two  inac- 
cessible objects. 

Measure  a  base  line  AB,  from 
whose  extremities  C  and  D  are 
visiblt!.  At  A  observe  the  angles 
CAD,  DAB ;  and  at  B  observe 
the  angles  CB A  and  CBD. 

Then,  in  the  triangle  ABC,  we  know  two  angles  and  th^ 
side  AB.  .-.  AC  may  be  found.  In  the  triangle  ABD  we 
know  two  angles  and  the  side  AB.     .*.  AD  may  be  found. 

Lastly,  in  the  triangle  ACD,  AC  and  AD  have  been  deter- 
mined, and  the  included  angle  CAD  has  been  measured ; 
and  thus  CD  can  be  found. 

EXAMPLES. 

1.  Let  AB  =  1000  yards,  the  angles  BAC,  BAD  =  76°  30' 
and  44°  10',  respectively ;  and  the  angles  ABD,  ABC  = 
81°  12'  and  40°  5',  respectively :  find  tlie  distance  between 
C  and  D.  .  ^ns.  669.8  yards. 


2.  A  and  B  are  two  trees  on  one  side  of  a  river ;  at  two 
stations  P  and  Q  on  the  other  side  observations  are  taken, 
and  it  is  found  that  the  angles  APB,  BPQ,  AQP  are  each 
equal  to  30°,  and  that  the  angle  AQB  is  equal  to  60°.  If 
PQ  =  a,  show  that 

AB  =  -V2l. 
6 

126.  The  Dip  of  the  Horizon.  —  Since  the  surface  of  the 
earth  is  spherical,  it  is  obvious  that  an  object  on  it  will  be 
visible  only  for  a  certain  distance  depending  on  its  height ; 
and,  conversely,  that  at  a  certain  lieight  above  the  ground 
the  visible  horizon  will  be  limited. 


THE  DIP   OF  THE  HORIZON. 


187 


Let  0  be  the  centre  of  the  eartli,  P  a  point  above  the  sur- 
face, PD  a  tangent  to  the  surface  at  « r— 71P 

D.  Then  D  is  a  point  on  the  terres- 
trial horizon ;  and  CPD,  which  is  the 
angle  of  depression  of  the  most  dis- 
tant point  on  the  horizon  seen  from 
P,  is  called  the  dip  of  the  horizon  at  P. 
The  angle  DOP  is  equal  to  it. 

Denote  the  angle  CPD  by  0,  the 
height  AP  by  h,  and  the  radius  OD 
by  r.    Then 

A  =  OP  -  OA  =  r  sec  e  —  r 


_  r  (1  -  cos  6] 

) 

cos^ 

h  cos  0 
1  —  cos  d 

' 

PD  =  r  tan  6  = 

1  — cos^ 

=  hcot^^   .     . 

(Art.  48,  Ex.  8) 

'  =  PAx  PB  =  /i(/i  +  2r) 

.     .     .     (Geom.) 

Also         ] 

Since,  in  all  cases  which  can  occur  in  practice,  h  is  very 
small  compared  with  2  r,  we  have  approximately 

PD'  =  2/ir. 

Let  n  •=  tlie  number  *  of  miles  in  PD,  Ji  =  the  feet  in  PA, 
and  r  =  4000  miles  nearly.     Then 


A=^  = 


2r 

5280^2 
8000 


(5280  n)^ 
8000  X  5280 

^  5.28  ^2  ^  2 


8 


W 


km 


i   I 


i  !i 


I'i'i  I 


m 


■;!•  J 


b  I! 


'f 


i'  I 


*  It  will  be  noticed  thnt  «  is  n  number  merely,  and  that  the  result  will  be  in  feel, 
Biace  the  luilea  have  been  reduced  to  feet. 


188 


PLANE  TRIGONOMETRY. 


That  is,  the  height  at  which  objects  can  be  seen  varies  as 
the  square  of  the  distance. 
Thus,  it  n  =  l  mile,  we  have 

/i  =  I  feet  =  8  inches  ; 

if  w  =  2  miles,  /i  =  -| .  2^  =  |  feet,  etc.,  etc. 

Thus  it  appears  that  an  object  less  than  8  inches  above 
the  surface  of  still  water  will  be  invisible  to  an  eye  on  the 
surface  at  the  distance  of  a  mile. 

Example.  From  a  balloon,  at  an  elevation  of  4  miles,  the 
dip  of  the  sea-horizon  is  observed  to  be  2°  33'  40":  find  (1) 
the  diameter  of  the  earth,  and  (2)  the  distance  of  the 
horizon  from  the  balloon. 

Alls.   (1)  8001.24  miles ;  (2)  178.944  miles. 

127.  Problem  of  Pothenot  or  of  Snellius.  —  To  determine 
a  point  in  the  plane  of  a  given  triangle,  at  which  the  sides  of 
the  triangle  subtend  given  angles. 

Let  ABC  be  the  given  triangle,  and 
P  the  required  point.  Join  P  with 
A,  B,  C. 

Let  the  given  angles  APC,  BPC  be 
denoted  by  a,  (3,  and  the  unknown 
angles  PAC,  PBC  by  x,  y  respectively ; 
then  a  and  ^  are  known  ;  and  when  x 
and  y  are  found,  the  position  of  P  can 
be  determined,  for  the  distances  PA 
and  PB  can  be  found  by  solving  the 
triangles  PAC,  PBC. 

We  have    a;-f?/  =  27r  —  a  —  /8  —  C 
6  sin  X  _  asin  y  __  pp 
sin  /8 


(1) 


Also 


sm  a 
Assume  an  auxiliary  angle  <^  such  that     ' 

.       ,      a.  sin  a 

tan  <^  =  —, ; 

b  sin/S 
then  the  value  of  ^  can  be  found  from  the  tables. 


EXAMPLES. 


189 


Thus, 


sin  X 
sin  y 


=  tan  <^» 


1.    Given     a =51.303, 

c  =  150; 

find      A=20°, 

B  =  70°, 

2.   Given     a  =  157.33, 

c=250; 

find     A  =  39°,        ... 

B=51°, 

sin  X  —  sin  ?/      tan  <A  —  1      ,       .  ,       .  ^o^ 

.-.  — r-^'  = ^ — --  =  tan  (^  —  45  ) 

sin  X  +  sin  y      tan  <f>  +  1 

[(14)  of  Art.  61]. 

.'.  tan  ^(x  —  y)  =  tan  ^{x-^-y)  tan  (c^  —  45°) 

[(13)  of  Art.  61]. 

=  tan  (45°  -  <^)  tan  L  (a  +  ;8  +  C)     .     (2) 
thus  from  (1)  and  (2)  x  and  y  are  found. 

EXAMPLES. 
Solve  the  following  right  triangles  : 

5       ■'■''-■■      '■-■..'. 
6  =  140.95. 

6=194.28. 

3.  Given     a=104,     .,,        c=185;  ,  ,       . 

find     A=34°12'19".6,  B  =  55°  47' 40 ".4,    6  =  153. 

4.  Given     a=304,  c=425; 

find      A=45°40'  2".3,  B=44°19'57".7,    6  =  297. 

6.    Given     6=3,  c=5; 

find      A=53°   7'48".4,  B=36°52'll".6,    a=4. 

6.  Given     6=15,  c=l7; 

find      A=28°   4'20".9,  B=61°55'39".l,    a=8.  - 

7.  Given     6=21,  -         c=29;  :      . 

find     A=43°36'10".l,  B  =  46°23'49".9,    a=20. 

8.  Given     6=7,  -;        c=25;      ^  *     ? 

find     A=73°44'23".3,  B  =  16°15'36".7,    a =24. 


< 


«!i 


'^m 


ifvi- 


190 


PLANE  TRIGONOMETRY. 


9. 

Given 

6=33, 

c=65; 

find 

A=oy°29'23".2, 

B  =  30°30'36".8, 

a --56. 

10. 

Given 

c=625, 

A=44°; 

find 

a =434.161, 

6=449.587. 

11. 

Given 

c=300, 

A=52''; 

find 

a =236.403, 

6=184.698. 

12. 

Given 

c=13, 

A=  67°22'48".5 

> 

find 

B=22°37'll".5, 

a=12 

6=5. 

13. 

Given 

A=77°19'10".6, 

c=41; 

\ 

find 

B  =  12°40'49".4, 

a=40, 

6=9. 

14. 

Given 

B=48°53'16".5, 

c=73; 

'  ,s  ■ 

find 

A=41°  6'43".5, 

a =48, 

6=55. 

16. 

Given 

B  =  64°  0'38".8, 

c=89; 

.       ■  ^,Ui';    S 

find 

A=25°59'21".2, 

a=39. 

6=80. 

16. 

Given 

A  =  77°19'10".6, 

a=40; 

find 

B  =  12''40'49".4, 

6=9, 

c=41. 

17. 

Given 

A=87°12'20".3, 

a=840; 

find 

B=  2°47'39".7, 

6=41, 

c=841. 

18. 

Given 

A=32°31'13".5, 

a=336; 

find 

B=57°28'46".5, 

6=527, 

c=625. 

19. 

Given 

A=82°41'44", 

a=1100; 

find 

B=  7°  18' 16", 

6=141, 

c=1109. 

20. 

Given 

A=75°23'18".5, 

6=195; 

'.  --\ 

find 

B  =  14°36'41".5, 

a=748, 

c=773. 

21. 

Given 

B  =  87°  49' 10", 

6=42536.37; 

find 

A=  2°  10' 50", 

a =1619.626, 

c=42567.2. 

EXAMPLES. 


191 


22.  Given  A: 

find    B: 

23.  Given  A: 

find    B: 

24.  Given  ct: 

find    A: 

25.  Given  a: 
'    find    A: 

26.  Given  a: 

find   A: 

27.  Given  a: 

find   A: 

28.  Given  a= 

find   A: 

29.  Given  a- 

find   A= 

30.  Given  «= 

find   A= 

31.  Given  a= 

find   A= 


88° 59' 
1°   V, 

35°  16' 25", 
54°43'3J  " 


6: 

a-, 
b-- 


-.  7694.5,  6: 

=  42°  15',  B. 

=  736,  ^         b: 

--  69°38'56".3,  B: 

=  200,  b: 

-.  18°  10' 50",     B: 

=276,         .         6: 
=  29°14'30".3,  B: 

:    396,  b: 

:  44°  29' 53",     B: 

=  278.3,  b: 

:     41°  30',  B: 

=372,  &= 

:  41°  16'  2". 7,  B  = 

:  526.2,  b: 

:  51°  45' 18". 7,  B  = 


=2.234875; 

=  125.9365,         'c=  125.9563! 

=388.2647; 

=548.9018,  c= 672.3412. 

=  8471;        ..■■  .V, 

=47°  45',  0=11444. 

=273; 

=20°  21 '3".  7,      c=785. 

:609; 

:  71°  49' 10",       c=641. 

=  493; 

J        ......  ...        I .  I 

:60°45'29."7,    c=565. 

=  403; 

:  45° 30'  7",     0=565. 

=  314.6;      . 

:  48°  30',  c=420. 

:423.924;      .  . 

:  48°43'57".3,  c=564. 

=  414.745; 

:  38°14'41".3,  c=670. 


Solve  the  following  oblique  triangles  : 

32.  Given  B=  50°  30',  C  =  122°  9', 

find   A=     7°  21', 

33.  Given  A=  82°20', 

find    C=  54°  20', 

34.  Given  A  =  79°  59', 


find    C=  51°  20', 


6=542.850, 

B=  43°  20', 
6=331.657, 

B=  44°  41', 
6=567.888, 


a =90; 
c= 595.638. 

a=479 ; 
c= 392.473. 

a=795; 
c= 663.986. 


% 


in 


liiiil ' 


'ill; 


192 


PLANE   I'RIGOIWMETRY. 


\ 


3.^. 

Given  B  = 

37°  58', 

C=  65°  2', 

a=999; 

find   A  = 

77°  0', 

&= 630.771, 

c= 829.480. 

36. 

Given  A  = 

70°  55', 

C=  52°  9', 

a=6412 ; 

find    B  = 

56°  56', 

6=5686.00, 

c= 5357.50. 

37. 

Given  A  = 

48°  20', 

B=  81°  2' 16", 

b=o.7o ; 

find    C  = 

50°  37' 44", 

a= 4.3485, 

c=4.5. 

38. 

Given  A  = 

72°   4', 

B=  41°  56' 18", 

c=24;   '■ 

find    C  = 

65°  59' 42", 

a  =  24.995, 

6=17.559. 

39. 

Given  A  = 

43°36'10".l, 

C  =  124°58'33".6 

,6  =  29;   •    .; 

find    B  = 

11°25'16".3, 

a  =  101, 

c=120. 

40. 

Given  A  = 

69°  59'  2".5, 

C=   70°  42' 30", 

6  =  149; 

find    B  = 

39°18'27".5, 

tt  =  221. 

c=222. 

41. 

Given  A  = 

21°  14' 25", 

«=345, 

6=695;     '■ 

find    B  = 

46°  52' 10", 

C  =  111°  53' 25", 

c= 883.65. 

or      B'= 

133°  7' 50", 

0'=  25°  37' 45", 

c'= 411.92. 

42. 

Given  A  = 

41°  13'   0", 

a =77.04, 

6  =  91.06; 

find   B  = 

51°  9'   6", 

C=  87°  37' 54", 

c= 116.82, 

.    or      B'= 

128°  50' 54", 

C'=     9°  56'   6", 

c'  =  20.172. 

43. 

Given  A= 

21°  14' 25", 

a =309, 

6=360; 

find    B  = 

24°  51' 54", 

C  =  133°  47' 41", 

c=615.67. 

or      B'= 

155°  L'  6", 

C'=     3°43'29'4 

c'= 55.41. 

44. 

Given  B  = 

68°  10' 24", 

a =83. 856, 

6=83.153; 

find    A= 

65°  5' 10", 

C=  45°  44' 26", 

c= 65.696. 

45. 

Given  B  = 

60°  0'32", 

a =27.548, 

6=35.055; 

find    A  = 

42°  53' 34", 

C=  77°   5' 54", 

c=. 39.453. 

46. 

Given  A  = 

60°, 

a=120, 

6  =  80; 

find    B  = 

35°  15' 52", 

C=  84°  44'  8", 

c=  137.9796 

EXAMPLES. 

198 

•1' 

47. 

Given 
find 

A=  50°,    . 
B=  38°  38 '24", 

a =119,    ,;   I 
C=  91°  21' 36", 

6  =  97;                              !; 
c=  155.3.                         ; 

48. 

Given 
find 
or 

C=  65°  59', 
A=  72°  4' 48", 
A' =107°  55' 12", 

rt=25, 

B=  41°  56' 12", 

B'=     6°  5' 48". 

cce24;                               1 

6=17.56,                          1 

1. 

i 

49. 

Given  A=  18°55'28".7, 
find    B=  67°22'48".l, 

a  =  13,                     6=37; 

or  B'  =  112°37'll".9.           .; 

50. 

Given 
find 

C=   15°11'21", 
A=  85°  11' 58", 

ci=232,  ' 

B=  79°  36' 40", 

6=229; 
c=61. 

51. 

Given 
find 

C  =  126°  12 '14", 
A=  32°  28' 19", 

rt=5132,      ^ 
B=  21°  19' 27", 

6=3476;    . 
c= 7713.3. 

52. 

Given 
find 

C=  55°  12'  3", 
A=  67°28'51".5, 

«  =  20.71, 

B=  57°19'5".5, 

6=18.87; 
c=  18.41. 

53. 

Given 
find 

C=  12° 35'  8", 
A =136°  15' 48", 

a  =8.54, 

B=  31°   9'  4", 

6=6.39; 
c=2.69.      - 

54. 

Given 
find 

C=  34°  9' 16", 
A=133°51'34", 

a =3184, 

B=  11°  59' 10", 

6=917; 
0=2479.2. 

55. 

Given 
find 

C=  32°10'53".8, 
A=136°23'49".9, 

a=101, 

B=  11°25'16".3, 

6=29; 

c=78.       ,":V^                1 

1,  , 

56. 

Given 
find 

C=  96°57'20".l, 
A,=  77°19'10".6, 

a=401, 

B=  5°43'29".2, 

6=41; 

c=408.                            j 

57. 

Given 
find 

C=  30°  40' 35", 
A=110°  0'57".5, 

a=221, 

B=  39°18'27".5, 

6  =  149;                            j 
c=120.                             i 

V   ■ 

58. 

Given 
find 

C=  66°59'25".4, 
A=  79°  36' 40",  . 

a =109,    ' 

B=  33°23'54".6, 

6=61; 
c=102. 

h   - 

1: 

1, 

69. 

Given 
find 

C=131°24'44", 
A=  33°23'54".6, 

* 

a =229,           V- 
B=  15°11'21".4, 

6=109; 

c=312.                             , 

v 

ii 

Mrikadi^itMkkMMM 


« 
w 


!•- 


194 


PLANE  THWONOMETJiY. 


60. 

Given  0=104"  3' 51", 

a=241, 

6=169; 

find  A =45°  46' 16". 

5,B  =  30°   9'52".5, 

c= 332.97. 

61. 

Given  a= 289, 

&=6()1, 

c=712; 

find  A=23''32'12", 

B=56°   8' 42", 

C=100°19'  6". 

62. 

Given  a =17, 

6=113, 

c=120; 

findA=  7°37'42", 

B =61°  55' 38", 

C  =  110°  26' 40". 

63. 

Given  a=  15.47, 

6=17.39, 

c= 22.88; 

find  A=42°30'44", 

B  =  49°  25' 49", 

C=88°3'27".     ': 

64. 

Given  a =5134, 

6=7268, 

c=9313; 

findA=33°15'39", 

B  =  50°56'  0", 

C=95°48'21". 

65. 

Given  a =99, 

6=101, 

c=158; 

find  A =37°  22' 19", 

B=38°15'41", 

C  =  104°22'0". 

66. 

Given  a=ll. 

6=13, 

c=16; 

findA=43°  2'56", 

B  =  53°  46' 44", 

C=83°10'20". 

67. 

Given  a =25, 

6=26, 

c=27; 

findA=56°15'  4", 

B=59°51'10", 

0  =  63°  53' 46". 

68. 

Given  a =197, 

6=53, 

c=240; 

find  A =31°  53' 26". 

8,B=  8°10'16".4,  C  =  139°56'16".8.           | 

69. 

Given  a  =  509, 

6=221, 

c=480;                          1 

find  A=84°32'50". 

5,  B=25°36'30".7,  C=69°50'38".8.            | 

70. 

Given  a =533, 

6=317, 

c  =  510; 

find  A =76°  18' 52", 

B=35°18'  0".9 

0=68°23'7".l. 

71. 

Given  a =565, 

6=445, 

c=606; 

find  A =62°  51' 32". 

9,B=44°29'53", 

0=72°38'34".l. 

72. 

Given  a =10, 

6=12, 

c=14; 

find  A=44°24'55". 

2,B=57°  7' 18", 

0  =  78°  27' 47". 

73. 

Given  a  =  .8706, 

6=.0916, 

c=.7902; 

find  A =149°  49' 0". 

4,B=  3°  1'56".2 

,0=27°9'3".4.                f 

EXAMPLES. 

Find  the  area : 

74.  Given  a=10,      6=12,       0  =  60". 


195 


Ana.  30V3. 


75. 

(( 

a =40, 

6=60,       C=30°.                             600. 

76. 

(( 

6=7, 

c=5V2,   A  =  135°.                           17^. 

77. 

u 

a=32.5, 

6=56.3,     C =47"  5' 30".                   670. 

78. 

i( 

6=149, 

A=   70° 42' 30",  B  =  39°  18' 28".   15540. 

79. 

(I 

c= 8.025, 

B  =  100°  5'23",  C=3r  6'12".  46.177. 

80. 

« 

a=5. 

6=6,         c=7.                                 12. 

81. 

« 

a =625, 

6=505,     c=904.                      151872. 

82. 

(( 

a =409, 

6=169,     c=510.                        30600. 

83. 

u 

a=577, 

6=73,       c=520.                      :  12480. 

84. 

il 

tt=52.53, 

6=48.76,  c=44.98.               1016.9487. 

86. 

« 

a =13, 

6=14,       c=15.         '                      84. 

86. 

({ 

a =242  yards,  6=1212  yards,  c=l450  yards. 

Ans.  6  acres. 

87. 

a 

a= 7.152, 

6=8.263,  c=9.375.                 28.47717. 

88.  The  sides  of  a  triangle  are  as  2  :  3  :  4 :  show  that  the 
radii  of  the  escribed  circles  are  as  ^ :  ^  :  1. 

89.  The  area  of  ^  triangle  is  an  acre ;  two  of  its  sides 
are  127  yards  and  150  yards :  find  the  angle  between  them. 

Ans.  30°  32' 23". 

90.  The  adjacent  sides  of  a  parallelogram  are  5  and  8, 
and  they  include  an  angle  of  60°:  find  (1)  the  two  diag- 
onals, and  (2)  the  area.        Ans.   (1)  7,  Vl29;    (2)  20  V3. 

91.  Two  angles  of  a  triangular  field  are  22^°  and  45°,  and 
the  length  of  the  side  opposite  the  latter  is  a  furlong.  Show 
that  the  field  contains  2^  acres. 


!l; 


196 


'\ 


PLANE  TRIGONOMETRY. 


HEIGHTS  AND  DISTANCES. 


92.  At  a  point  200  feet  in  a  horizontal  line  from  the  foot 
of  a  tower,  the  angle  of  elevation  of  the  top  of  the  tower  is 
observed  to  be  60° :  find  the  height  of  the  tower. 

Ans.  346  feet. 

93.  From  the  top  of  a  vertical  cliff,  the  angle  of  depres- 
sion of  a  point  on  the  shore  150  feet  from  the  base  of  the 
cliff,  is  observed  to  be  30° :  find  the  height  of  the  cliff. 

Ans.  86.6  feet. 

94.  From  the  top  of  a  tower  117  feet  high,  the  angle  of 
depression  of  the  top  of  a  house  37  feet  high  is  observed  to 
be  30° :  how  far  is  the  top  of  the  house  from  the  tower  ? 

Alls.  138.6  feet. 

95.  The  shadow  of  a  tower  in  the  sunlight  is  observed 
to  be  100  feet  long,  and  at  the  same  time  the  shadow  of  a 
lamp-post  9  feet  high  is  observed  to  be  3 V3  feet  long :  find 
the  angle  of  elevation  of  the  sun,  and  height  of  the  tower. 

Ans.  60°;   173.2  feet. 

96.  A  flagstaff  25  feet  high  stands  on  the  top  of  a 
house;  from  a  point  on  the  plain  on  which  the  house 
stands,  the  angles  of  elevation  of  the  top  and  bottom  of 
the  flagstaff  are  observed  to  be  60°  and  45°  respectively : 
find  the  height  of  the  house  above  the  point  of  observation. 

Ans.  34.15  feet. 

97.  From  the  top  of  a  cliff  100  feet  high,  the  angles  of 
depression  of  two  ships  at  sea  are  observed  to  be  45°  and 
30°  respectively  ;  if  the  line  joining  the  ships  points  directly 
to  the  foot  of  the  cliff,  find  the  distance  between  the  ships. 

Ans.  73.2. 

98.  A  tower  100  feet  high  stands  on  the  top  of  a  cliff ; 
from  a  point  on  the  sand  at  the  foot  of  the  cliff  the  angles 


EXAMPLES. 


197 


jlitf; 


of  elevation  of  the  top  and  bottom  of  the  tower  are  observed 
to  be  75°  and  GO"  respectively :  find  the  height  of  the  cliff. 

Ans.  8(5.6  feet. 

99.  A  man  walking  a  straight  road  observes  at  one  mile- 
stone a  house  in  a  direction  making  an  angle  of  30°  with 
the  road,  and  at  the  next  milestone  the  angle  is  60° :  how 
far  is  the  house  from  the  road  ?  Ans.  1524  yds. 

100.  A  man  stands  at  a  point  A  on  the  bank  AB  of  a 
straight  river  and  observes  that  the  line  joining  A  to  a  post 
C  on  the  opposite  bank  makes  with  AB  an  angle  of  30°. 
He  then  goes  400  yards  along  the  bank  to  B  and  finds  that 
BC  makes  with  BA  an  angle  of  60°:  find  the  breadth  of 
the  river.  '  Ans.  173.2  yards. 

101.  From  the  top  of  a  hill  tlse  angles  of  depression  of 
the  top  and  bottom  of  a  flagstaff  25  feet  high  at  the  foot 
of  the  hill  are  observed  to  be  45°  13'  and  47°  12'  respectively  : 
find  the  height  of  the  hill.  Ans.  373  feet. 

102.  From  each  of  two  stations,  east  and  west  of  each 
other,  the  altitude  of  a  balloon  is  observed  to  be  45°,  and 
its  bearings  to  be  respectively  N.W.  and  N.E. ;  if  the  sta- 
tions be  1  mile  apart,  find  the  height  of  the  balloon. 

Ans.  3733  feet. 

103.  The  angle  of  elevation  of  a  balloon  from  a  station 
due  south  of  it  is  60°,  and  from  another  station  due  west 
of  the  former  and  distant  a  mile  from  it  is  45° :  find  the 
height  of  the  balloon.  Ans.  6468  feet. 

104.  Find  the  height  of  a  hill,  the  angle  of  elevation  at 
its  foot  being  60°,  and  at  a  point  500  yards  from  the  foot 
along  a  horizontal  plane  30°.  A^is.  250V3  yards. 

105.  A  tower  51  feet  high  has  a  mark  at  a  height  of  25 
feet  from  the  ground :  find  at  what  distance  from  the  foot 
the  two  parts  subtend  equal  angles.       :  ,  :,•    .^-.  ^.'.  ,::..«  Jy 


V 


11; 


f' 


'ji'< 


If';  ■^: 


198 


PLANE ,  TRIGONOMETRY. 


106.  The  angles  of  a  triangle  are  as  1 :  2  :  3,  and  the  per- 
pendicular from  the  greatest  angle  on  the  opposite  side  is 
30  yards  :  find  the  sides.  ;  .4ns.  20 V3,  60,  40 V3. 

107.  At  two  points  A,  B,  an  object  DE,  situated  in  the 
same  vertical  line  CE,  subtends  the  same  angle  a ;  if  AC, 
BC  be  in  the  same  right  line,  and  equal  to  a  and  b,  respec- 
tively, prove  ;  ..    ,  ■  i         .i 

DE  =  (a  +  ?j)  tan  «. 

108.  From  a  station  B  at  the  foot  of  an  inclined  plane 
BC  the  angle  of  elevation  of  the  summit  A  of  a  mountain  is 
60°,  the  inclination  of  BC  is  30°,  the  angle  BOA  135°,  £uid 
the  length  of  BC  is  1000  yards  :  find  the  height  of  A  over  B. 

1    ;;,.j  '  ^HS.  500(3 -f-V3)  yards. 

109.  A  right  triangle  rests  on  its  hypotenuse,  the  length 
of  which  is  100  feet;  one  of  tlie  angles  is  30°,  and  the 
inclination  of  the  plane  of  the  triangle  to  the  horizon  is 
60°:    find  the  height  of  the  vertex  above  the  ground. 

Ans.  2oV3cos  18°. 

110.  A  station  at  A  is  due  west  of  a  railway  train  at  B  ; 
after  traveling  N.W.  6  miles,  the  bearing  of  A  -from  the 
train  is  S.  22|-°  W. :  find  the  distance  AB.         Ans.  6  miles. 

111.  The  angles  of  depression  of  the  top  and  bottom  of  a 
column  observed  from  a  tower  108  feet  high  are  30°  and  60° 
respectively:  find  the  height  of  the  column.      Ans.  72  feet. 

112.  At  the  foot  of  a  mountain  the  elevation  of  its  sum- 
mit is  found  to  be  45°.  After  ascending  for  one  mile,  at  a 
slope  of  15°,  towards  the  summit,  its  elevation  is  found  to 
be  60° :  find  the  height  of  the  mountain. 

y3-fi 

V2 


Ans. 


miles. 


113.  A  and  B  are  two  stations  on  a  hillside.  The  inclina- 
tion of  the  hill  to  the  horizon  is  30°.  The  distance  between 
A  and  B  is  500  yards.     C  is  the  summit  of  another  hill  in 


EXAMPLES. 


199 


the  same  vertical  plane  as  A  and  B,  on  a  level  with  A,  but 
at  B  its  elevation  above  the  horizon  is  15° :  find  the  distance 
between  A  and  C.  Ans.  500(\/,S4-l). 

114.  From  the  top  of  a  cliff  the  angles  of  depression  of 
the  top  and  bottom  of  a  lighthouse  97.25  feet  high  are 
observed  to  be  2,3°  17'  and  24°  19'  respectively  :  how  much 
higher  is  the  cliff  than  the  lighthouse  ?  Ans.  1942  feet. 

■  115.  The  angle  of  elevation  of  a  balloon  from  a  station 
due  south  of  it  is  47°  18'  30",  and  from  another  station  due 
west  of  the  former,  and  distant  C71.38  feet  from  it,  the 
elevation  is  41°  14' :  find  the  height  of  the  balloon. 

Ans.  1000  feet. 

116.  A  person  standing  on  the  bank  of  a  river  observes 
the  elevation  of  the  top  of  a  tree  on  the  opposite  bank  to  be 
51° ;  and  when  he  retires  80  feet  from  the  river's  bank  he 
observes  the  elevation  to  be  46° :  find  the  breadth  of  the 
river.  .  Ans.  155.823  feet. 

117.  From  the  top  of  a  hill  I  observe  two  milestones  on 
the  level  ground  in  a  straight  line  before  me,  and  I  find 
their  angles  of  depression  to  be  respectively  5°  and  15° : 
find  the  height  of  the  hill.  Ans.  228.6307  yards. 

118.  A  tower  is  situated  on  the  top  of  a  hill  whose  angle 
of  inclination  to  the  horizon  is  30°.  The  angle  subtended 
by  the  tower  at  the  foot  of  the  hill  is  found  by  an  observer 
to  be  15°;  and  on  ascending  485  feet  up  the  hill  the  tower 
is  found  to  subtend  an  angle  of  30° :  find  (1)  the  height  of 
the  tower,  and  (2)  the  distance  of  its  base  from  the  foot  of 
the  hill.  Ans.   (1)280.015;   (2)  705.015  feet. 

119.  The  angle  of  elevation  of  a  tower  at  a  place  A  due 
south  of  it  is  30° ;  and  at  a  place  B,  due  west  of  A,  and  at 
a  distance  a  from  it,  the  elevation  is  18° :   show  that  the 

height  of  the  tower  is 


I 


•\/2  +  2  VO 


200 


PLANE  TRIGONOMETRY. 


I'l: 


,1  '( 


120.  On  the  bank  of  a  river  there  is  a  cohimn  200  feet 
high  supporting  a  statue  30  feet  high.  The  statue  to  an 
observer  on  the  opposite  bank  subtends  an  equal  angle  with 
a  man  6  feet  high  standing  at  the  base  of  the  column:  find 
the  breadth  of  the  river.  Ans.  10V115  feet. 

121.  A  man  walking  along  a  straight  road  at  the  rate  of 
3  miles  an  hour,  sees  in  front  of  him,  at  an  elevation  of  60°, 
a  balloon  which  is  travelling  horizontally  in  the  same  direc- 
tion at  the  rate  of  6  miles  an  hour ;  ten  minutes  after  he 
observes  that  the  elevation  is  30° :  prove  that  the  height  of 
the  balloon  above  the  road  is  440  V3  yards. 

122.  An  observer  in  a  balloon  observes  the  angle  of 
depression  of  an  object  on  the  ground,  due  south,  to  be 
35°  30'.  The  balloon  drifts  due  east,  at  the  same  elevation, 
for  2^  miles,  when  the  angle  of  depression  of  the  same 
object  is  observed  to  be  23°  14' :  find  the  height  of  the 
balloon.  Ans.  1.34394  miles. 

123.  A  column,  on  a  pedestal  20  feet  high,  subtends  an 
angle  45°  to  a  person  on  the  ground ;  on  approaching  20 
feet,  it  again  subtends  an  angle  45° :  show  that  the  height 
of  the  column  is  100  feet. 

124.  A  tower  51  feet  high  has  a  mark  25  feet  from  the 
ground :  find  at  what  distance  the  two  parts  subtend  equal 
angles  to  an  eye  5  feet  from  the  ground.  Ans.  IGO  feet. 

125.  From  the  extremities  of  a  sea-wall,  300  feet  long, 
the  bearings  of  a  boat  at  sea  were  observed  to  be  N.  23°  30' 
E.,  and  N.  35°  15'  W. :  find  the  distance  of  the  boat  from 
the  sea-wall.  Ana,  262.82  feet. 

126.  ABC  is  a  triangle  on  a  horizontal  plane,  on  which 
stands  a  tower  CD,  whose  elevation  at  A  is  50°  3'  2" ;  AB  is 
100.62  feet,  and  BC  and  AC  make  with  AB  angles  40°  35'  17" 
and  9°  59'  50"  respectively  :  find  CD.         Ans.  101.166  feet. 


EXAMPLES. 


201 


127.  The  angle  of  elevation  of  a  tower  at  a  distance  of 
20  yards  from  its  foot  is  three  times  as  great  as  the  angle 
of  elevation  100  yards  from  the  same  point :  show  that  the 

height  of  the  tower  is  "^ —  feet. 

128.  A  man  standing  at  a  point  A,  due  south  of  a  tower 
built  on  a  horizontal  plain,  observes  the  altitude  of  the  tower 
to  be  60°.  He  then  walks  to  a  point  B  due  west  from  A 
and  observes  the  altitude  to  be  45°,  and  then  at  the  point  C 
in  AB  produced  he  observes  the  altitude  to  be  30° :  prove 
that  AB  =  BC. 

129.  The  angle  of  elevation  of  a  balloon,  which  is  ascend- 
ing uniformly  and  vertically,  when  it  is  one  mile  high  is 
observed  to  be  35°  20' ;  20  minutes  later  the  elevation  is 
observed  to  be  55°  40' :  how  fast  is  the  balloon  moving  ? 

Ans.  3  (sin  20°  20')  (sec  55°  40')  (cosec  35°  20')  miles  per  hour. 

130.  The  angle  of  elevation  of  the  top  of  a  steeple  at  a 
place  due  south  of  it  is  45°,  and  at  another  place  due  west 
of  the  former  station  and  distant  100  feet  from  it  the 
elevation  is  15° :  show  that  the  height  of  the  steeple  is 
50(3^ -3-i)  feet. 

131.  A  tower  stands  at  the  foot  of  an  inclined  plane 
whose  inclination  to  the  horizon  is  9°;  a  line  is  measured 
up  the  incline  from  the  foot  of  the  tower,  of  100  feet  in 
length.  At  the  upper  extremity  of  this  line  the  tower  sub- 
tends an  angle  of  54° :  find  the  height  of  the  tower. 

Ans.  114.4  feet. 

132.  The  altitude  of  a  certain  rock  is  observed  to  be  47°, 
and  after  walking  1000  feet  towards  the  rock,  up  a  slope 
inclined  at  an  angle  of  32°  to  the  horizon,  the  observer  finds 
that  the  altitude  is  77° :  prove  that  the  vertical  height  of 
the  rock  above  the  first  point  of  observation  is  1034  feet. 


202 


PLANE  TRIGONOMETBY. 


133.  From  a  window  it  is  obfserved  that  the  angle  of 
elevation  of  the  top  of  a  house  on  the  opposite  side  of  the 
street  is  29°,  and  the  angle  of  depression  of  the  bottom  of 
the  house  is  56°:  find  the  height  of  the  house,  supposing 
the  breadth  of  the  street  to  be  80  feet.        Aiis.  162.95  feet. 


134.  A  and  B  are  two  positions  on  opposite  sides  of  a 
mountain ;  C  is  a  point  visible  from  A  and  B ;  AC  and  BC 
are  10  miles  and  8  miles  respectively,  and  the  angle  BCA  is 
60°:  prove  that  the  distance  between  A  and  B  is  9.165 
miles. 

135.  P  and  Q  are  two  inaccessible  objects  ;  a  straight  line 
AB,  in  the  same  plane  as  P  and  Q,  is  measured,  and  found 
to  be  280  yards ;  the  angle  PAB  is  95°,  the  angle  QAB  is 
47^°,  the  angle  QBA  is  110°,  and  the  angle  PBA  is  52°  20' : 
find  the  length  of  PQ.  Ans.  509.77  yards. 

136.  Two  hills  each  264  feet  high  are  Just  visible  from 
each  other  over  the  sea :  how  far  are  they  apart  ?  (Take 
the  radius  of  the  earth  =  4000  miles.)  Ans.  40  miles. 

137.  A  ship  sailing  out  of  harbor  is  watched  by  an 
observer  from  the  shore ;  and  at  the  instant  she  disappears 
below  the  horizon  he  ascends  to  a  height  of  20  feet,  and 
thus  keeps  her  in  sight  40  minutes  longer :  find  the  rate  at 
which  the  ship  is  sailing,  assuming  the  earth's  radius  to  be 
4000  miles,  and  neglecting  the  height  of  the  observer. 

Alls.  40V330  feet  per  minute. 

138.  From  the  top  of  the  mast  of  a  ship  64  feet  above 
the  level  of  the  sea  the  light  of  a  distant  lighthouse  is  just 
seen  in  the  horizon ;  and  after  the  ship  has  sailed  directly 
towards  the  light  for  30  minutes  it  is  seen  from  the  deck 
of  the  ship,  which  is  16  feet  above  the  sea :  find  the  rate 
at  which  the  ship  is  sailing.     (Take  radius  =  4000  miles.) 

Ans.  8V||  miles  per  hour. 


EXAMPLES. 


203 


139.  A,  B,  C,  are  three  objects  at  known  distances  apart ; 
namely,  AB  =  1056  yards,  AC  =  924  yards,  BC  =  1716 
yards.  An  observer  places  himself  at  a  station  P  from 
which  C  appears  directly  in  front  of  A,  and  observes  the 
angle  CPB  to  be  14°  24' :  find  the  distance  CB. 

Ans.  2109.824  yards. 

140.  A,  B,  C,  are  three  objects  such  that  AB=o20  yards, 
AC  =  600  yards,  and  BC  =  435  yards.  Brom  a  station  B 
it  is  observed  that  ABB  =  15°,  and  BBC  =  30° :  find  the 
distances  of  P  from  A,  B,  and  C ;  the  point  B  being  near- 
est to  B,  and  the  angle  ABC  being  the  sum  of  the  angles 
ABB  and  BBC.  Ans.  BA  =  777,  BB  =  502,  BC  =  790. 


204 


PLANE  TRIGONOMETRY. 


CHAPTER  VIII. 

OONSTRUOTION  OF  LOGAKITHMIO  AND  TEIGONOMETEIO 

TABLES. 


128.  Logarithmic  and  Trigonometric  Tables.  —  In  Chap- 
ters IV.,  v.,  and  VII.,  it  was  shown  how  to  use  logarithmic 
and  trigonometric  tables;  it  will  now  be  shown  how  to 
calculate  such  tables.  Although  the  trigonometric  func- 
tions are  seldom  capable  of  being  expressed  exactly,  yet 
they  can  be  found  approximately  for  any  angle ;  and  the 
calculations  may  be  carried  to  any  assigned  degree  of  accu- 
racy. We  shall  first  show  how  to  calculate  logarithmic 
tables,  and  shall  repeat  here  substantially  Arts.  208,  209, 
210,  from  the  College  Algebra. 

129.  Exponential  Series.  —  To  expand  e'  in  a  series  of 
ascendimj  powers  ofx. 

By  the  Binomial  Theorem, 


/i   1   ly       11        1  ,  nx(nx-l)  1 

\       nj  n  12         n^ 


Similarly, 

1\" 


[2 
nx{nx  —  l){nx  —  2)  1    . 

xltc )      x{x — \{x ) 


13 


(1) 


1  + 


n . 


i_i  A-i-Yi-'' 


1  +  1  + 


n 


12 


+ 


n, 


n 


13 


+ 


(2) 


LOGARITHMIC  SERIES. 


206 


and  therefore  series  (1)  is  equal  to  series  (2)  however  great 
n  may  be.  Hence  if  n  be  indefinitely  increased,  we  have 
from  (1)  and  (2) 

1  _(_  a;  +  -  +  ^  +  ^ 4- ...  = /^l  4- 1  4- i -f- i  4- 1  +  ...Y. 
The  series  in  the  parenthesis  is  usually  denoted  by  e; 


hence 


x'  .  x' 


x^ 


e'=l+x  +  ^  +  '^  +  ^^-^ 


(3) 


[2  '  [3  ■  |4 

which  is  the  expansion  of  e'^  in  powers  of  x. 
This  result  is  called  the  Exponential  Theorem. 

If  we  put  03=  1,  wo  have  from  (3)  ,        ,  _, 

•'■■:'.  6  =  1  +  1  ^  1  +  1  +  i  4-  i  +  1  +  ... 

From  this  series  we  may  readily  compute  the  approxi- 
mate value  of  e  to  any  required  degree  of  accuracy.  This 
constant  value  e  is  called  the  Napierian  base  (Art.  64).  To 
ten  places  of  decimals  it  is  found  to  be  2.7182818284. 

Cor.  Let  a  =  e'' ;  then  c  =  log^  a,  and  a'  =  e".  Substi- 
tuting in  (3),  we  have 


c-af 


c^a^ 


e-=l  +  ca;4-^+^  + 


or 


a' 


=  J  +  0,  log,  a  +  ^Q|^%  ?^5|iO!  + 


(4) 


which  is  the  expansion  of  a""  in  powers  of  x. 


130.  Logarithmic   Series. —  To  expand  loge(l4-a;)  in  a 
series  of  ascending  powers  of  X.  '    ■        <_    •      ,- 

By  the  Binomial  Theorem,  •       ' 

a'  =  (l  +  a-l)'  =  l  +  a;(a-l)  +  "^^-^~^)(a-l)^ 

+  ^^-^)(-^'--)(a-ir+..- 


206 


PLANE  TRIGONOMETRY. 


+  terms  involving  x-,  x^,  etc. 

Comparing  this  value  of  a'  with  that  given  in  (4)  of  Art. 
129,  and  equating  the  coefficients  of  x,  we  have 

Put  a  =  I  -f  iB ;  then 

••••••     (3) 


log.(l4-«)  =  a;--  +  - 


a' 


+ 


This  is  the  Logarithmic  Series;  but  unless  x  be  very 
small,  the  terms  diminish  so  slowly  that  a  large  number  of 
them  will  have  to  be  taken  ;  and  hence  the  series  is  of  little 
practical  use  for  numerical  calcul  ition.  If  a;>  1,  the  series 
is  altogether  unsuitable.  We  shall  therefore  deduce  some 
more  convenient  formulae. 


Changing  x  into  —  x,  (1)  becomes 


X- 


ar' 


cfi" 


log.(l  —  x)  =  —  x 

Subtracting  (2)  from  (1),  we  have 


log, 


'  —   /  (  1!  _L I I |_ 


Put 


l~x 

l-\-x      n-\-l 


I{  XA h 

\  -'-5        5 


X  = 


7 
1 


(2) 


(3) 


2w  +  l' 


1  —X         n 
and  (3)  becomes 

W  !i±l  =  2r      ^       4- ^ + ^ + 

^'     n  l2n  +  l      3{2n-\-iy      o{2n-[-iy 

or  log„  (n  4- 1) 


=log.n+2r-l 
|_2?H 


.+?:, 


-.+■ 


(4) 


i  +  l     3(2n  +  l)3     r)(2»  +  l)« 

This  series  is  rapidly  convergent,  and  gives  the  logarithm 
of  either  of  two  consecutive  numbers  to  any  extent  when 
the  logarithm  of  the  other  number  is  known. 


COMPUTATION   OF  LOGARITHMS. 


207 


131.  Computation  of  Logarithms.  —  Logarithms  to  the 
base  e  are  called  Napierian  Logarithms  (Art.  64).  They 
are  also  called  natural  logarithms,  because  they  are  the 
first  logarithms  which  occur  in  the  investigation  of  a 
method  of  calculating  logarithms.  Logarithms  to  the  base 
10  are  called  commoyi  logarithms.  When  logarithms  are 
used  in  theoretical  investigations,  the  base  e  is  always 
understood,  just  as  in  all  practical  calculations  the  base 
10  is  invariably  employed.  It  is  only  necessary  to  com- 
pute the  logarithms  of  prim,e  numbers  from  the  series, 
since  the  logarithm  of  a  composite  number  may  be  obtained 
by  adding  together  the  logarithms  of  its  component  factors. 
The  logarithm  of  1  =  0.  Putting  w  =  1,  2,  4,  6,  etc.,  suc- 
cessively, in  (4)  of  Art.  130,  we  obtain  the  following 


log,.  2  =  2 


Napierian  Logarithms : 
■1.1.1    +J_H-J-  + 


log.  3=log,2-f-2 
log.  4  =  21og.2 
log.  5=  log.  4 +2 


3     3 .  S"^    5 . 3«     7 .  3^     9 . 3' 
1.1.1,1 


5    3.5'    5.5^    7-r/ 


9    3 .  93    5 . 9-^    7  •  9^ 
log.  6  =  log.2  +  log.3  ,,  - 

13     3 .  13^^     5 .  13* 


+ 


+ 


log.  7  =  log/, +  2 

log.  8  =  31og.2 
log.  9  =  21og,3 
log.l0  =  log,r)4-log.2 
And  so  on. 


=  0.69314718. 

=  1.09861228. 

=  1.38629436. 
=  1.60943790. 

=  1.79175946. 

=  1.94590996. 

=  2.07944154. 
=  2.19722456. 
=  2.30258509. 


The  number  of  terms  of  the  series  which  it  is  necessary 
to  include  diminishes  as  n  increases.     Thus,  in  computing 


208 


PLANE  TRIGONOMETRY. 


the  logarithm  of   101,  the  first  term  ol    the  series  gives 
the  result  true  to  seven  decimal  places. 

By  changing  b  to  10  and  a  to  e  in  (1)  of  Art.  65,  we  have 

or         common  log  m  =  Napierian  logm  x  .43429448. 

The  n timber  .43429448  is  called  the  modulus  of  the  common 
system.     It  is  usually  denoted  by  fi. 

Hence,  the  common  logarithm  of  any  number  is  equal  to 
the  Napierian  logarithm  of  the  same  number  multiplied  by 
the  modulus  of  the  common  system,  .43429448. 

Multiplying  (4)  of  Art.  130  by  fi,  we  obtain  a  series  by 
which  common  logarithms  may  be  computed;  thus, 

logio(ri  +  l)  = 

Common  Logarithms. 
Iog,o2  =  fi  log,  2  =  .43429448  x    .09314718  =  .3010300. 
Iogio3  =  /^log,3  =  .43429448  x  1.09861228  =  .4771213. 
log,o  4  =  2  logio  2  =  .6020600. 

log,„5  =  /ilogeO  =  .43429448  x  1.60943790  =  .6989700. 
And  so  on. 

132.  If  $  be  the  Circular  Measure  of  an  Acute  Angle, 
sin  6,  $,  and  tan  9  are  in  Ascending  Order  of  Magnitude. 

With  centre  O,  and  any  radius,  de- 
scribe an  arc  BAB'.  Bisect  the  angle 
BOB'  by  OA;  join  BB',  and  draw  the 
tangents  BT,  B'T. 

Let  AOB  =  AOB'  =  6.     Then 

BB'  <  arc  BAB'  <  BT  +  B'T 

(Geom.,  Art.  246) 
.-.  EC  <  arc  BA  <  BT. 


LIMITING    VALUES  OF  SIN 0. 


209 


BC      BA      BT 
•  ■  OB      OB      OB' 

'  .-.    sin  ^  <  6/ <  tcau  d. 

133.  The  Limit  of  ^^,  when  6  is  Indefinitely  Diminished, 
is  Unity.  .  - 

We  have       sin  d<0<U\id (Art.  132) 

e 


1< 


sin  6 


sec^. 


C) 


Now  as  d  is  diminished  indefinitely,  sec  $  approaches  the 

limit  unity;  then  when  ^  =  0,  we  have  sec^  =  1. 

B 
.'.  the  limit  of ,  which  lies  betwt:n  seed  and  unity, 

IS  unity.  ^        ^    ■  •  . 

.'.  also  — -—  approaches  the  limit  unity. 

6  .-I  ',» 

A     tau  d     sin  6           ^  ,-,     ■,-    ■,     n  tan  d      ,        /\  ■     • 
As  = X  secP,  the  limit  of  ,  when  6  is  in- 

0  0  0 

definitely  diminished,  is  also  unity. 

This  is  often  stated  briefly  thus : 

sin  0     ^         -,  tan  0     i       x^      ^      n. 
=  1,  and  — — -  =  1,  when  0  =  0. 


6 


6 


Note.  —  From  this  it  follows  that  the  sines  and  the  tangents  of  very  small 
angles  are  proportional  to  the  angles  themselves. 


134.  If  0  is  the  Circular  Measure  of  an  Acute  Angle,  sin  0 

ffi                                             ffi 
lies  between  0  and  0  —  —  \  and  cos  0  lies  between  1 and 

^      0'      0*  * 

2      16 


(1)  We  have 


^00 
tan  -  >  - 

2     2 


.00       $ 
sin  -  >  -  cos  -• 

2     2       2 


(Art.  132) 


210  PLANE  TliWONOMETliY. 

.'.  2  sin  -  COS  -  >  ^  COS"* -• 
2       2  2 

>/l_^       .     .     (Art.  132) 

.-.  BmB<e  and  >^-?- 

4 

(2)  cos^=l-2sin2-. 

ffi 
.'.  cos^>l--. 

2  ,.   „ 

Also,  sin|>^-^(|)^y(l).       '^      , 

...eos.<l-2||-|] 

2      16      512 

.-.  cose>l-?-and  <1_.^  +  ^. 
2  2      lb 

* 

Note.  —It  may  be  proved  that  sin  fl>  S ,  as  follows : 

6 

We  have  3  sin  -  -  sin  0  =  4  sin^  -  (Art.  50) (1) 

3  3 

.'.  3  sin  -  -  sin  ?  =  1  sin'  --  (by  putting  -  for  0) (2) 

3*  3  3'  3 

3sini--sln-^  =4ein^  ?- (n) 

3»  3"-^  S" 

Multiply  (1),  (2), ...  (n)  by  1,  3, ...  S"-!,  respectively,  and  add  them, 

3"  sin  -  -  sin  9  =  4^8in3  -  +  3  sin^  -  +  -  3"-l  sin^  -V 
3"  \        3  3»  3V 


SINE  AND  COSINE  OF  lO"  AND  OF  l\ 


211 


.    (1) 

.     (2) 
.    (n) 


.to* 


...*._ .,n»<4(3^  +  -  +  ...^2,^) (Art 


«• 


132) 


33      V        3»  ^in-2) 

tin  — 
3" 
If  n  B  oe ,  then  — ;: —  =  1 

4^,/,      1  1    \     4*<       1         #a 

and       3,9a(l  +  ^  +  ...^^_J  =  — ._  =  -^-. 


'.' » 


(Art.  133) 


3» 


.♦.  «-sln9<-,  and  .-.  •ln«>»-??. 
6  6 


Thii  makeH  tho  limltB  for  sin  0  closer  than  In  (1)  of  this  Art. 

135.  Te  calculate  the  Sine  and  Cosine  of  10"  and  of  1 '. 

(1)  Let  0  be  the  circular  measure  of  10".  ' 


Then 


e  = 


10 1 


180  X  60  X  60 


3.141592653589793 
64800 


or         ^  =  .000048481368110  •  •  -,  correct  to  16  decimal  places. 


=  .000000000000032...,       « 

...  e_i'=. 000048481368078...,      « 
4 


(( 


(( 


u 


<( 


It 


(I 


Hence  the  two  quantities  6  and  6 agree  to  12  deci- 

mal  places ;  and  since  sin B <d  and  > ^  —  -7  (Art.  134), 

.-.  sin  10"  =  .000048481368,  to  12  decimal  places. 
We  have 


cos  10"  =  Vl-sinno"  =  1  -  i  sin2 10" 

=  .9999999988248  .••,  to  13  decimal  places. 

Or  we  may  use  the  results  established  in  (2)  of  Art.  134, 
and  obtain  the  same  value. 


:^i 


212 


PLANE  TRIGONOMETRY. 


(2)  Let  6  be  the  circular  measure  of  1'. 
Then    '  ^  •   '      '  "'* 


49  = 


ir 


-  =  .000290888208665,  to  15  decimal  places. 


(( 


180  X  60 

.-.-  =  .000000000006        to  12        « 

-■'■-  4 

.-.  tf--  =  . 00029088820  toll        «  « 

4  ■.      /    '■' 

Hence  6  and  0  —  —  differ  only  in  the  twelfth  decimal. 
,-.  sin  1'  =  .00029088820  to  11  decimal  places. 


cos  1'  =  VT-sin21'=. 999999957692025  to  15  decimal  places, 

Otherwise  iiius : 

1  -  ^'  =.999999957692025029  to  18  decimal  places. 
2  ^ 

and      —  =  .00000000000000044  to  17  decimal  places. 
16 

Butcosl'>l-^';-nd<l-^  +  :^  .     .     .     (Art.  134) 

.'.  cos  1'  =  .9999999576920.5  to  15  decimal  places,  as  before. 

Cor.  1.  The  sine  of  10"  equals  the  circular  measure  of  10", 
to  12  decimal  places  ;  and  the  sine  of  V  equals  the  circular 
measure  ofVto  11  decimal  places.  ' 

Cor.  2.  Ifn  denote  any  number  of  seconds  less  than  60,  we 
shall  have  ajyproximately 

sin  n"  =  n  sin  1", 

for  the  sine  of  n"  =  the  circular  measure  of  n",  approxi- 
mately, =  n  times  the  circular  measure  of  1". 

n      Q            circular  measure  of  n"  .      ,  ,        ,,    , 

Cor.  3.   n  = '"TTi '   approximately ;   that 

is,  the   number  of  seconds  in  any  small  angle  is   found 


;   that 
found 


TABLES  OP  NATURAL  SINES  AND  COSINES,     213 

approximately  by  dividing  the   circular   measure  of  that 
angle  by  the  sine  of  one  second.       •>.    .      •>  ■■  '^ 

136.  To  construct  a  Table  of  Natural  Sines  and  Cosines  at 
Intervals  of  1'. 

We  have,  by  Art.  45,  '  •■     .  "    ' 

sin {x  +  y)  =  2  sin  x  cos  y  —  sin (x  —  y),         •'■■■'■ 

GOs{x -\- y)  =  2Gosxcosy —  c.os{x —  y).  'm-.'- 

Suppose  the  angles  to  increase  by  1';  putting  3/ =  1',  we 
have,     ;•,..•>    I •'   :  ;,.,•■'  :''',■■..,,.,..,';:,;  /j 

jv     sin(aj4- 1')  =  2sina;cosl'  — sin(aj  — 1')  .     .     .     .     (I) 

cos(.T  + 1')  =  2cos.'Bcos  1' ~  cos(a;  — 1')  .     ...     (2) 

Putting  a;=  1',  2',  rV,  4',  etc.,  in  (1)  and  (2),  we  get  for 

the  sines 

sin  2'  =  2sin  1'  cos  1'  -  sin  0'  =  .0005817764,         ' ' 

sin  3' =  2  sin  2' cos  1' -  sin  1' =  .0008726646, 

sin  4'  =  2  sin  3'  cos  1'  -  sin  2'  =  .0011635526 ; 

and  for  the  cosines 

cos  2'  =  2  cos  1'  cos  1'  -  cos  0'  =  .9999998308, 
cos  3'  =  2  cos  2'  cos  1'  -  cos  1 '  =  .9999996193, 
cos  4' =  2  cos  3' cos  1' -  cos  2' =  .9999993223.     '    .'^ 

We  can  proceed  in  this  manner*  until  we  find  the  values 
of  the  sines  and  cosines  of  all  angles  at  intervals  of  1'  from 

0°to30°.  ,    ,  .  .=     ...(      ,:,. 

137.  Another  Method. 

Let  a  denote  any  angle.    Then,  in  the  identity, 

sin  (n  -f- 1)«  =  2  sin  na  cos  a  —  sin  {n  —  l)a, 
put  2(1  —  cos  «)  =  k, 

and  we  get 

sin(w4-l)«— sin  ?)«=sin  7J«  — sin(?j  — 1)«  — A;  sinn« 

*  This  method  is  due  to  Thomua  Siiupsou,  au  Eugliitb  geometriciaa. 


(1) 


'■  i-' 


b 


i\ 


!i 


214 


PLANE  TRIGONOMETRY. 


II 


li?    It 


This  formula  enables  us  to  construct  a  table  of  sines  of 
angles  whose  common  difference  is  a. 

Thus,  suppose  «  =  10",  and  let  n  =  1,  2,  3,  4,  etc. 

Then 

sin  20"  -  sin  10"  =  sin  10"  -  A;  sin  10", 

sin  30"  -  sin  20"  =  sin  20"  -  sin  10"  -  k  sin  20", 

sin  40"  -  sin  30"  =  sin  30"  -  sin  20"  -  Tc  sin  30",  etc. 

These  equations  give  in  succession  sin  20",  sin  30",  etc. 

It  will  be  seen  that  the  most  laborious  part  of  this  work  is 

the  multiplication  of  k  by  the  sines  of  10",  20",  etc.,  as  they 

are  successively  found.     But  from  the  value  of  cos  10",  we 

have 

A;  =  2  (1  -  cos  10")  =  .0000000023504, 

the  smallness  of  which  facilitates  the  process.  • 

In  the  same  manner  a  table  of  cosines  can  be  constructed 
by  means  of  the  formula, 

cos  (w  +  l)a  —  cos  na  =  cos  na  —  cos  (ii  —  l)a  —  k  cos  na, 

which  is  obtained  from  the  identity, 

cos  {n-{-l)a  =  2  cos  na  cos  a  —  cos  (?t  —  l)a, 

by  putting  2(1  —  cos  u)  =  k,  as  before. 

138.  The  Sines  and  Cosines  from  30°  to  G0°.  —  It  is  not 
necessary  to  calculate  in  this  way  the  sines  and  cosines  of 
angles  beyond  30°,  as  we  can  obtain  their  values  for  angles 
from  30°  to  60°  more  easily  by  means  of  the  formulae  (Art. 
45): 

sin  (30°  +  «)  =  cos  «  -  sin  (30°  -  rt), 

cos  (30°  +  «)  =  cos  (30°  -  n)  -  sin  «, 

by  giving  a  all  values  up  to  30°.     Thus, 

sin  30°  1' =  cos  1' -  sin  29°  59', 

cos  30°  1'  =  cos  29°  59'  -  sin  1',  and  so  on. 


TABLES  OF  TANGENTS  AND  SECANTS. 


215 


139.  Sines  of  Angles  greater  than  45".  —  When  the  sines 
of  angles  up  to  45°  have  been  calculated,  those  of  angles 
between  45°  and  90°  may  be  deduced  by  the  formula 

sin  (45°  +  a)  -  sin  (45°  -  «)  =  V^  sin  a       .     (Art.  45) 

Also,  when  the  sines  of  angles  up  to  60°  have  been  found, 
the  remainder  up  to  90°  can  be  found  still  more  easily  from 
the  formula 

sin  (60°  +  «)  -  sin  (G0°  -  a)  =  sin  «. 

Having  com])leted  a  table  of  sines,  the  cosines  are  known, 

since  •    .^ao        \ 

cos  a  =  sin  (90  —  a) . 

Otherwise  thus :  When  the  sines  and  cosines  of  the  angles 
up  to  45°  have  been  obtained,  those  of  angles  between  45° 
and  90°  are  obtained  from  the  fact  that  the  sine  of  an 
angle  is  equal  to  the  cosine  of  its  complement,  so  that  it  is 
not  necessary  to  proceed  in  the  calculation  beyond  45°. 

Note.  —  A  more  modern  method  of  calculating  the  sines  and  cosines  of  angles 
is  to  use  series  (3)  and  (4)  of  Art.  156. 

140.  Tables  of  Tangents  and  Secants.  —  To  form  a  table 
of  tangents,  we  find  the  tangents  of  angles  up  to  45°,  from 
the  tables  of  sines  and  cosines,  by  means  of  the  formula 


tan  a  = 


sin  a 
cos  a 


Then  the  tangents  of  angles  from  45°  to  90°  may  be 
obtained  by  means  of  the  identity  * 

tan  (45°  4-  «)  =  tan  (45°  -  a)  +  2  tan  2  a. 

When  the  tangents  have  been  found,  the  cotangents  are 
known,  since  the  cotangent  of  any  angle  is  equal  to  the 
tangent  of  its  complement. 

A  table  of  cosecants  may  be  obtained  by  calculating  the 
reciprocals  of  the  sines ;   or  they  may   be  obtained  more 

"^  Called  Cagnoli's  formula. 


HI 


f  \-i 


216 


PLANE  TRIGONOMETRY. 


easily  from  the  tables  of  the  tangents   by  means  of  the 

formula  c: 

coser- a  =  tan  - -}- cot  a. 

The  secants  are  then  known,  since  the  secant  of  any 
angle  is  equal  to  the  cosecant  of  its  complement. 

141.  FormulsB  of  Verification. —  Formulce  used  to  test  the 
accuracy  of  the  calculated  sines  or  cosines  of  angles  are  called 
Formulce  of  Verification. 

It  is  necessary  to  have  methods  of  verifying  from  time 
to  time  the  correctness  of  the  values  of  the  sines  and 
cosines  of  angles  calculated  by  the  preceding  method,  since 
any  error  made  in  obtaining  the  value  of  one  of  the  func- 
tions would  be  repeated  to  the  end  of  the  work.  For  this 
purpose  we  may  compare  the  value  of  the  sine  of  any  angle 
obtained  by  the  preceding  method  with  its  value  obtained 
independently.    -"■  ■         '  < 

•\/5  —  1 
Thus,  for  example,  we  know  that  sin  18°  =  — (Art. 

4 
57)  ;  hence  the  sine  of  18°  may  be  calculated  to  any  degree 
of  approximation,  and  by  comparison  with  the  value  obtained 
in  the  tables,  we  can  judge  how  far  we  can  rely  upon  the 
tables. 

Similarly,  we  may  compare  our  results  for  the  angles 
22^°,  30°,  36°,  45°,  etc.,  calculated  by  the  preceding  method 
with  the  sines  and  cosines  of  the  same  angles  as  obtained 
in  Arts.  26,  27,  56,  57,  58,  etc. 

There  are,  however,  certain  well-known  formulae  of  veri- 
fication which  can  be  used  to  verify  any  part  of  the  calcu- 
lated tables :  these  are 

Eider^s  Formulce:  .,  .        ,, 

sin  (36°  +  A)  -  sin  (36*'  -  A)  +  sin  (72°  -  A)  .,    ; 

,  .       i,i,j.    M    .    -  sin  (72° -f  A)  =  sin  A.       / 
cos  (36°  +  A)  +  cos  (36°  -  A)  -  cos  (7'>° -f  A) 

,  .   ,      —  cos  (<2°  — A)  =  cos  A. 


LOGARITHMIC  TEIGONOMETlilC  FUNCTIONS.    217 


Legendre's  Formula: 

sin  (54°  +  A)  +  sin  (54°  -  A)  -  sin  (18°  +  A) 

•"'-:;o  :■   i.;;    ..  #s.kH  1^:^^«      -  sin  (18°  -  A)  =  cos  A.  til 

The  verification  consists  in  giving  to  A  any  value,  and 
taking  from  the  tables  the  sines  and  cosines  of  the  angles 
involved:  these  values  must  satisfy  the  above  equations. 


To  prove  Euler's  Formuloi : 
sin  (36°  +  A)  -  sin  (36°  -  A)  =  2  cos  36°  sin  A 

VS  +  l      •         A 

=  -^ — - —  sin  A 
2 

sin  (72°  +  A)  -  sin  (72°  -  A)  =  2  cos  72°  sin  A 


2 


sin  A 


(Art.  45) 
(Art.  58) 

(Art.  45) 
(Art.  57) 


Subtracting  the  latter  from  the  former,  we  get  sin  A. 
Similarly,  Euler's  second  formula  may  be  proved. 
By  substituting  90°  —  A  for  A  in  this  formula  we  obtain 
Legendre's  Formula. 

142.  Tables  of  Logarithmic  Trigonometric  Functions.  — 

To  save  the  trouble  of  referring  twice  to  tables  —  first  to 
the  table  of  natural  functions  for  the  value  of  the  function, 
and  then  to  a  table  of  logarithms  for  the  logarithm  of  that 
function  —  it  is  convenient  to  calculate  the  logarithms  of 
trigonometric  functions,  and  arrange  them  in  tables,  called 
tables  of  logarithmic  sines,  cosines,  etc. 

When  tables  of  natural  sines  and  cosines  have  been  con- 
structed, tables  of  logarithmic  sines  and  cosines  may  be 
made  by  means  of  tables  of  ordinary  logarithms,  which  will 
give  the  logarithm  of  the  calculated  numerical  value  of  the 
sine  or  cosine  of  any  angle ;  adding  10  to  the  logarithm  so 
found  we  have  the  corresponding  tabular  logarithm.  The 
logarithmic  tangents  may  be  found  by  the  relation 

log  tan  A  =  10  4-  log  sin  A  —  log  cos  A ;  • 

and  thus  a  table  of  logarithmic  tangents  may  be  constructed. 


MV^ 


218 


PLANE  TRIGONOMETRY. 


PROPORTIONAL  PARTS. 

143.  The  Principle  of  Proportional  Parts.  —  It  is  often 
necessary  to  find  from  a  table  of  logarithms,  the  logarithm 
of  a  number  containing  more  digits  than  are  given  in  the 
table.  In  order  to  do  this,  we  assumed,  in  Chapter  IV.,  the 
principle  of  proportional  parts,  which  is  as  follows  : 

The  differences  between  three  ntimhers  are  proportional  to  the 
corresponding  differences  between  their  logarithms,  provided 
the  differences  between  the  numbers  are  small  compared  with 
the  numbers. 

By  means  of  this  principle,  we  are  enabled  to  use  tables 
of  a  more  moderate  size  than  would  otherwise  be  necessary. 

We  shall  now  investigate  how  far,  and  with  what  excep- 
tions, the  principle  or  rule  of  proportional  increase  is  true. 


!l>l 


I 


144.  To  prove  the  Rule  for  the  Table  of  Common  Loga- 
rithms. 

Wp  have  .  '  }■■. 

log  (n  4- rf)  -  logn  =  log^t_^  =  log  f  1  + '^ 

where  /u,  =  .43429448  •••,  a  quantity  <  i- 

Now  let  n  be  9,n  integer  not  <  10000,  and  d  not  >  1 ; 

then  ,     -  is  not  greater  than  .0001.  .•         " 

n,,      ^^^^-^  ^ 

.'.  ;^  is  not  >  4(-0001)^  i.e.,  not  >  .0000000025: 

2n^       ,.  . ...  .  ,...,...     ;,.  .      -   ' 

and        ^- is  much  less  than  this. 
'3n^  ; 

.-.  log  (n  +  rf)  —  log  n  =  fx  ,  correct  at  least  as  far  as  seven 
decimal  places.  ^      ■        '* 


RULE  OF  PROPORTIONAL   PARTS. 


219 


Hence  if  the   number  be  changed  from  n  to  n  +  d,  the 

corresponding  change   in  the  logarithm  is  approximately 
fid 
n 
Therefore,  the  change  of  the  logarithm  is  approximately 

proportional  to  the  change  of  the  number. 

145.  To  prove  the  Rule  for  the  Table  of  Natural  Sines. 

sin  {6  +  h)  —  sin  6  =  sin  h  cos  6  —  sin  6  (1  —  cos  h) 

=  sin /i  cos  6(1  —  tan  ^  tan  -  I  .    (Art.  51) 

If  h  is  the  circular  measure  of  a  very  small  angle,  sin  h  =  h 
nearly,  and  tan  -  =  -  nearly. 

,:    .*.  sin  {0  +  h)  —  sin  d  =  h  cos  6(1  —  tan  B  tan  -  j 

=  7icos6 sin  6. 

2 

If  h  is  the  circular  measure  of  an  angle  not  >  1',  then 

h  is  not  >  .0003  (Art.  135).  .-.  ^'  is  not  >  .00000005;  and 
sin  6  is  not  >  1. 

.-.  sin(6+/0  —sin  d=h  cos  $,  as  far  as  seven  decimal  places, 
which  proves  the  ]>roposition. 

Similarly,  sin  {6  —  h)  —  sin  d  =  —  h  cos  6,  approximately. 

146.  To  prove  the  Eule  for  a  Table  of  Natural  Cosines. 

cos  {6  —  h)  —  cos  6  =  sin  h  sin  6  ~  cos  6(1  —  cos  h) 

h\ 


sin  h  sin  6(1  —  cot  6  tan 


2. 


If  h  is  the  circular  measure  of  a  very  small  angle,  sin  h  —  h 
nearly,  and  tan -  =  - nearly. 

;      .'.  cos  (6  — 7i)  — cos  6— /i  sin 6(1 —  cot 6  tan  ^  J 

=  i'isin6—  -cos  6. 


I  , 


i      « 


220 


PLANE  TRIGONOMETRY. 


We  may  prove,  as  in  Art.  145,  that 

-  cos  0  is  not  >  .00000005. 

.'.  cos(0  —  7i)  —  GosO  =  f  sind,  as   far  as  seven  decimal 
places,  which  proves  the  proposition. 

Similarly,  cos  (^  +  7i)-- cos  d  =  — /i  sin^,  approximately. 

147.  To  prove  the  £11]  e  for  a  Table  of  Natural  Tangents. 


tan  {d  +  h)  -  tan  0  =  «i'L(^„+J)  _  ^lil^  = 


sin  h 


cos  {6  +  h)      cos  0     cos  {0  -\-  h)  cos  d 

,      ,  __  tan  A 

cos*^  ^  (1  —  tan  0  tan /t) 

/  '     "  ■■ 

If  h   is   the   circular   measure   of   a   very   small   angle, 
tan /i  =  A  nearly. 

h  sec''  0 


tan  {0  +  h)  —  tan  d  = 


1  —  h  tan  6 


:  i\!^t'i'f\^ 


=  h  sec-  6  +  h^  sin  $  sec^  $. 


.'.  tan  (^  -f-  h)  —  tan  ^  =  7i  sec^  6,  approximately, 

unless  sin  $  sec^^  is  large,  which  proves  the  proposition,  n  .•• 
Similarly,  cot  {6  —  h)  —  cot  6  =  h  cosec"  $,  approximately. 

Sch.  1.  If  h  is  the  circular  measure  of  an  angle  not  >  1', 
then  h  is  not  >  .0003.     Hence  the  greatest  value  of  h^  sin  0 

sec"'  0  is  not  >  .00000009  sin  6  sec'^  $.    Therefore,  when  d>'^, 

4 

we  are  liable  to  an  error  in  the  seventh  place  of  decimals. 
Hence  the  rule  is  not  true  for  tables  of  tangents  calculated 
for  every  minute,  when  the  angle  is  between  45°  and  90°.      , 

Sch.  2.  Since  the  cotangent  of  an  angle  is  equal  to  the 
tangent  of  its  complement,  it  follows  immediately  that  the 
rule  must  not  be  used  for  a  table  of  cotangents,  calculated 
for  every  minute,  when  the  angle  lies  between  0°  and  45°. 


BULE  OF  PROPORTIONAL  PARTS. 


221 


rr 


148.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Sines. 


7j2 

sin  {6  +  h)  —  sin  0  =  h  oos  9  -  j-  sin  6 

sin  ^  2 

.'.  log  sin  (0  4- ^)  —  log  sin  ^ 

=  ,*log^l  +  /tcot0-|'^ 


(Art.  145) 


=  al  ^  cot  B  —   --  —-(hcot 


A  ' 


D'-] 


(Art.  130) 


f.h\ 


= /x/i  cot  ^  -  ^  (1  +  cot^  ^)  +  . . . 

=  fihcotd  —  ^cosec-0-\ 

If  h  is  the  circular  measure  of  an  angle  not  >  10",  then 
h  is  not  >  .00005,  and  therefore,  unless  cot  0  is  small  or 
cosec^  0  large,  we  have 

log  sin  (^  +  ^)  —  log  sin  ^  = /i/i  cot  ^, 

as  far  as  seven  decimal  places,  which  proves  the  rule  to  be 
generally  true. 

Sch.  1.  When  6  is  small,  cosec  6  is  large.  If  the  log 
sines  are  calculated  to  every  10",  then  h  is  not  >. 00005, 
and  fi  is  not  >  .5. 


.-.  ^fxh^  cosec^O  is  not  > 


6  cosec^  6 


10 


10 


fmt 


In  order  that  this  error  may  not  affect  the  seventh  decimal 
place,  ecosec'^^  must  not  be  >  10^,  that  is,  6  must  not  be 
less  than  about  5°.  ^  .  >:  , 

When  $  is  small,  cot  0  is  large.     Hence,  when  the  angles 


fin 

8 


222 


PLANE  TRIGONOMETRY, 


are  small,  the  differences  of  consecutive  log  sines  are  irregu- 
lar, and  they  are  not  insensible.  Therefore  the  rule  does 
not  apply  to  the  log  sine  when  the  angle  is  less  than  5°. 

Sch.  2.  When  6  is  nearly  a  right  angle,  cot^  is  small, 
and  cosec  6  approaches  unity. 

Hence,  when  the  angles  are  nearly  right  angles,  the  dif- 
ferences of  consecutive  log  sines  are  irregular  and  nearly 
insensible. 


149.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Co- 
sines. 

cos(^  —  h)  —  cos  6  =  h  sin  6  —  —cos  0  .     (Art.  146) 

.       ...  cos(^-/0^i^;,tan^-^. 
cos^  2 

.•.  logcos(0  — /i)  —  logcosd 

/u.  log  (  1  + /i  tan  ^ ] 


=  J  I 


ht^inO  -  —  -Uhtan  0  -  ^Y 


= /x/itan  ^  —  ^-sec^^-f  •••  •  '<  ',■  - 

In  this  case  the  differences  will  be  irregular  and  large 
when  0  is  nearly  a  right  angle,  and  irregular  and  insensible 
when  B  is  nearly  zero.  This  is  also  clear  because  the  sine 
of  an  angle  is  the  cosine  of  its  complen^ent. 

150.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Tan- 
gents. 

tan  {B  +  li)  -  tan B=h  sec' Q  -f  W sin e  sec"' B  .     (Art.  147) 


tan  (B  -\-  h) 
t&nB 


1  +  -. 


h 


sin  B  cos  $ 


+  /i'sec2^. 


BULE  OF  PliOPOliTlONAL  PARTS. 


223 


log  tan  {6  4-  h)  —  log  tan  6 


—H- 


sin  0  cos  $ 


+h:^sec^e- 


1/        Ji 


2Vsin^cose 

= t!:^ +  aJi'fseG^e 

sin  ^  cos  ^'^    V  2sin='^cos2tf 

.-.  log  tan  {6  4-  h)  —  log  tan  6 


■{-hHec'e]  +•• 


h 


■] 


'»• 


sin  6  cos  $ 


sm"2e 


151.  Cases  where  the  Principle  of  Proportional  Parts  is 
Inapplicable. 

It  appears  from  the  last  six  Articles  that  if  h  is  small 
enough,  the  differences  are  proportional  to  h,  for  values  of  6 
which  are  neither  very  small  nor  nearly  equal  to  a  right  angle. 

The  following  exceptional  eases  arise  : 

(1)  The  difference  sin  {$  +  '0  —  sin  6  is  insensible  when 
6  is  nearly  90°,  for  in  that  case  h  cos  6  is  very  small ;  it  is 
then  also  irregular ^  for  ^W&mB  may  become  comparable 
with  h  cos  0. 

(2)  The  difference  cos  {S  -\-  h)  —  cos  $  is  both  insensible 
and  irregular  when  6  is  small. 

(3)  The  difference  tan  (d  -\-  h)  —  tan  6  is  irregular  when 
6  is  nearly  90°,  for  hJ^  sin  6  sec^  ^  may  then  become  compar- 
able with  h  sec*^  ^ ;  it  is  never  insensible,  since  sec  6  is  not 
<1.  .,.,,,,,,;-,..,,,^_  ,,.,,,._,..     .,.,„3.,.:v  .;...:: 

(4)  The  difference  log  sin  {$  +  h)  —  log  sin  6  is  irregular 
when  0  is  small,  and  both  irregular  and  insensible  when  ^  is 
nearly  90°.  ^ 

(5)  The  difference  log  cos  (0  +  /a)  —  log  cos  ^  is  insensible 
and  irregular  when  ^  is  small,  and  irregular  when  ^  is 
nearly  90°.  , 

(6)  The  difference  log  tan  (^  -f  ^)  —  log  tan  ^  is  irregular 
when  0  is  either  small  or  nearly  90°.      ..  ,  „     ,, 


224 


PLANE  TRIGONOMETliV. 


m 


A  difference  which  is  insensible  is  also  irregular;  but  the 
converse  does  not  hold. 

When  the  differences  for  a  function  are  insensible  to  the 
number  of  decimal  places  of  the  tables,  the  tables  will  give 
the /mw ''flows  when  the  angle  is  known,  but  we  cannot  use 
the  tables  to  find  any  intermediate  angle  by  means  of  this 
function;  thus,  we  cannot  determine  6  from  the  value  log 
cos  0,  for  small  angles,  or  from  the  value  log  sin  6,  for  angles 
nearly  90°. 

Wlien  the  differences  for  a  function  are  irregular  without 
being  insensible,  the  approximate  method  of  proportional 
parts  is  not  sufficient  for  the  determination  of  the  angle  by 
means  of  the  function,  nor  the  function  by  means  of  the 
angle ;  thus,  the  approximation  is  inadmissible  for  log  sin  0, 
when  6  is  small,  for  log  cos  6,  when  6  is  nearly  90°,  and 
for  log  tan  6  in  either  case.     (Compare  Art.  81.) 

In  these  cases  of  irregularity  without  insensibility,  the 
following  three  means  may  be  used  to  effect  the  purpose  of 
finding  the  angle  corresponding  to  a  given  value  of  the 
function,  or  of  the  function  corresponding  to  a  given  angle.* 


152.  Three  N^thods  to  replace  the  Rule  of  Proportional 

Parts.  ,v'  j-^7     '--;   '-i   T '11  ,v,   \■;^■.  ,•■;,/.  ;;,,,  ■;•''; 

(1)  The  simplest  plan  is  to  have  tables  of  log  sines  and 
log  tangents,  for  each  second,  for  the  lirst  few  degrees  of 
the  quadrant,  and  of  log  cosines  and  log  cotangents,  for 
each  second,  for  the  few  degrees  near  90°.  Such  tables  are 
generally  given  in  trigonometric  tables  of  seven  places ; 
we  can  then  use  the  principle  of  proportional  parts  for  all 
angles  which  are  not  extremely  near  0°  or  90°. 

(2)  Delambre's  Method.  In  this  method  a  table  is  con- 
structed which  gives  the  value  of  log [- log  sin  1"  for 

every  second  for  the  first  few  degrees  of  the  quadrant. 


*  This  article  has  been  taken  substantially  from  Hobson's  Trigonometry. 


rii 


METHODS   TO  REPLACE  THE  RULE. 


225 


it; 


Let  B  1)6  the  circular  measure  of  n  seconds.     Then,  when 
6  is  small,  we  have  6  =  nsin  1",  approximately. 


8in<> 


smw 


.•.  log  — — -  =  log — r— 7-,  =  log  sin  n"  —  log  n  —  log  sin  1". 
$  nsinl" 

.-.  log  sin  n"  =  log  n  +  Aog-~  +  log  sin  1"Y 

Hence,  if  the  angle  is  known,  the  table  gives  the  value 
of  the  expression  in  parenthesis,  and  log  n  can  be  found 
from  the  ordinary  table  of  the  logs  of  numbers ;  thus  log 
sin  rt"  can  be  found. 

If  log  sin  n"  is  given,  we  can  find  approximately  the  value 
of  n,  and  then  from  the  ta^^le  we  have  the  value  of  the 
expression  in  parenthesis ;  thus  we  can  find  log  w,  and  then 
n  from  an  ordinary  table  of  logs  of  numbers. 

Rem.   When  $  is  small  (less  than  5°), 


sin^      ^      ^  •      1.  1 
=  I  — -,  approximately 

$  6 


(Art.  134,  Note) 


Hence  a  small  error  in  0  will  not  produce  a  sensible  error 

in  the  result,  since  log  ^-— -  will  vary  much  less  rapidly 
than  e.  ^ 

(3)  Maskelyne^s  Method.  The  principle  of  this  method  is 
the  same  as  that  of  Delambre's.  If  ^  is  a  small  angle,  we 
have  '.         . 


sin  tf  =  0  —  — ,  approximately, 
6 


'i  f 


and 


cos^=l— — , 


sin^ 


i 


u 


u 


(( 


(Art.  134) 


=  (cos^)* 
log  sin  $=  log  0  +  \  log  cos  0,  approximately. 


Ml 


226 


PLANE  riilGONOMETRY. 


When  ^  is  a  small  angle,  the  differences  of  log  cos  9  are 
insensible  (Art.  149) ;  hence,  if  6  be  given,  we  can  find 
log^  accurately  from  the  table  of  natural  logarithms,  and 
a,lso  an  approximate  value  of  log  cos  6 ;  the  formula  then 
gives  log  sin  0  at  once. 

If  log  sin  6  be  given,  we  must  first  find  an  approximate 
value  of  B  from  the  table,  and  use  that  for  finding  log  cos  6, 
approximately ;  0  is  then  obtained  from  the  formula. 


IMS' 


EXAMPLES. 


1.  Prove  1  +  2 +,-  +  ,-+ •••  =  2e. 

2.  Prove  log^  =  l-/^ — ? —  +  — ^ — +  ... 

^2      2     U  •3.2'''     2.5.2* 


3.    Prove 


e.^\      1+2      14-2  +  3 

2     1^        [3  [4 


4.  Prove        1  =  ,^ +  i  +  ,^+ ... 

e     [3     [5     [7 

5.  Prove  tan  0  +  ^  tan'' ^  +  ^  tan' ^  + 


=  ilog| 


.os(.-l)^ 


(' 


,cos (  B  + 


IT 


6.  Prove  logell  =  2.39789527  •.•,  by  (4)  of  Art.  130. 

7.  Prove  log.  13  =  2.56494935..-,        «  « 

8.  Prove  log,  17  =  2.83321334...,        "  « 


9.   Prove  log,  19  =  2.9444394 


(( 


(( 


10.  Find,  by  means  of  the  table  of  common  logarithms 
and  the  modulus,  the  Napierian  logarithms  of  1325.07, 
62.9381,  and  .086623.  Ans.  7.18923,  3.96913,  -  2.4578. 


EXAMPLES. 


227 


11.   Prove  that  the  limit  of  m  sin  —  is  0,       when  m  =  oo. 

m 


12. 

<( 

13. 

« 

14. 

ii 

16. 

U 

16. 

(( 

17. 

(( 

18, 

(( 

19. 


20. 


21. 


t( 


a 


u 


u 


ii 


» 


(I 


<( 


n 


if 


a 


u 


"        m  tan  —  is  9, 
m 


u 


(( 


ii 


u 


—  sin  —  IS  7rt 

2  n 


,.2  U 


Trr^  tan  -  is  tt)^, 
n 

versa^  .    a'' 
vers  bO       6^' 

I  cos  -  )  IS  1, 
V      nj 

( Sin  .-  )  IS  1, 

[  COS  -  )    IS  1, 


sin  -\ 
n 


e 


n 


is  1, 


cos  -  I    IS  e    2j 


"   (rff 


ii 


m  =  30. 


n  =  oc. 


n  =  30. 


«       ^  =  0. 


« 


n  =  QO. 


n  =  QO. 


"  Jl  =  00. 


n  =  00. 


"      « — 


n  =  00. 


is  zero,  when  7i  =  oo. 


22.   If  0  is  the  circular  measure  of  an  acute  angle,  prove 
(l)cosd<l-^  +  f',  and  (2)  tan^>^  +  |- 


23.    Given  ^  =  ^:  prove  that  ^  =  4°  24',  nearly. 


24.   Given 


d        1014 

sing  ^2165. 
$        2166" 


prove  that  6  =  3°,  nearly. 


228 


PLANE  TRIGONOMETRY. 


!!  i-  i  ' 


25.   Given  sin  «/>  =  71  sin  6,  tan  <f}  =  2  tan  6 :  find  the  limit- 
ing values  of  n  that  these  equations  may  coexist. 

Ans.  n  must  lie  between  1  and  2,  or  between  —  1  and  —  2. 


26.   Find  the  limit  of  (cos  a»)  •=""=•='*',  when  a;  =  0. 


Ans. 


2ft2 


27.  From  a  table  of  natural  tangents  of  seven  decimal 
places,  show  that  when  an  angle  is  near  60°  it  may  be 
determined  within  about  ^-^  of  a  second. 

28.  When  an  angle  is  very  near  64°  36',  show  that  the 
angle  can  be  deiormined  from  its  log  sine  within  about  ^^ 
of  a  second ;  having  given  (log^  10)  tan  64°  36'  =  4.8492,  and 
the  tables  reading  to  seven  decimal  places. 


I 


DE  MOIVRE'8   THEOBEM. 


229 


\ 


CHAPTER  IX. 

DE  MOIVRE'S  THEOEEM.*  — APPLIOATIONS. 

153.  De  Moivre's  Theorem.  — For  any  value  ofn,  'positive 
or  negative,  integral  or  fractional. 

(cos  e  4-  V^^  sin  &) "  =  cos  nd  +  V^^  sin  nS  .     .     (1) 

I.    Wlien  n  is  a  ^wsitive  integer. 
We  have  the  product 


(cos  a  +  V-  1  sin  a)  (cos  /8  +  V-  1  sin  ;3) 

=  (cosacos/3— sinasinj8)  +  V^(cos«sin/3+sinacos^) 

=  cos  («  +  /8)  +  V^^  sin  («  +  /3) . 
Similarly,  the  product 
[cos  (a  +  /3)  +  V^T  sin  («  +  /S)  ]  [cos  y  +  V^3  sin  y] 
=  cos  («  4-  i3  +  y)  +  V^^  sin  (a  +  /3  +  y). 

Proceeding  in  this  way  we  find  that  the  product  of  any 
number  n  of  factors,  each  of  the  form 

cos  a  +  V—  1  sin  a  =  cos  («  +  ^  +  y  H n  terms) 


^■. 


4-V— lsin(«H-^  +  y  H ?i  terms). 

•     Suppose  now  that  «  =  /3  =  y  =  etc,  =  6,  then  we  have 

(oos  B  +  V^^  sin  6)"  =  cos  nO  +  V  —  1  sin nd, 
which  proves  the  theorem  when  n  is  a  positive  integer. 

♦  From  the  name  of  the  French  geometer  who  diecovered  it. 


280 


PLANE  TRIGONOMETRY. 


^^^^hF  ' 

I^^^^^HiJ  - 

HHr ''''' 

';;«! 

^^ 

!:*' 


it 


I 


II.    Wlien  n  is  a  negative  integer. 

Let  n  =  —  m ;  then  m  is  a  positive  integer.     Then 

(cos  e  +  V^^  sin  ey  =  (cos  6  +  V^l  sin  6)-"" 

^ 1  ^ 1 

(cosd  +  V— 1  sin^)*"      cosm^  +  V— 1  sinmd 

cos  md  —  V  —  i  sin  mB 


(by  I.) 


X 


COS  md  +  V  —  1  sin  md     cos  mQ  —  V  —  1  sin  m^ 


cosm^  —  V— 1  sinm^  a        / — r   •        n 

— — - — r—n =  cos  w^  —  V  —  1  sin  md 

cos'*  md  +  sur  wi6> 


=  cos  (  —  md)  +  V  —  1  sin  (  —  m6) . 

.'.  (cosd4- V— 1  sin^)"  =  cosw^+V— 1  sin  n^, 
which  proves  the  theorem  when  w  is  a  negative  integer. 

III.    When  n  is  a  fraction,  positive  or  negative. 
Let  w  =  -,  where  p  and  q  are  integers.     Then 

(cos^-HV^sin^)''=cosjj^+V^sinp^(by  Land  II.). 
But  (cos-^  +  V— isin-^j  =cos|)^+V-lsinpe. 

.-.   (cos  e  +  V-1  sin  ey  =  f  cos  -^  +  V^^  sin -^^' 
.-.   (cos  e  -\-  V^^  sin  ey  =  cos^^  +  V^  sin^^ ; 


that  is,  one  of  the  values  of  (cos  ^  +  V—  1  sin  ^)» 


IS 


pd 


pd 


cos 1-  V—  1  sin  — 

9  9 


In  like  manner, 

(costf  —  V—  1  sin  dy  =  cos  nd  —  V—  Isinn^. 

Thus,  De  Moivre's  Theorem  is  completely  established. 
It  shows  that  to  raise  the  binomial  cos^-f  V— 1  sin  tf  to 


I 


DE  MOIVRE'S   THEOREM. 


231 


any  power,  we  have  only  to  multiply  the  arc  0  by  the 
exponent  of  the  power.  This  theorem  is  a  fundamental 
one  in  Analytic  Mathematics. 


154.  To  find  All  the  Values  of  (cos^  +  V- Ising)^".— 
When  n  is  an  integer,  the  expression  (c()s^  + V—  Isin^)" 

can  have  only  one  value.     But  if  n  is  a  fraction  =  -,  the 
expression  becomes 

(cos  ^  +  V^^sin  ey=yj (con  ^  +  V^^  sin  d)", 

which  has  q  different  values,  from  the  principle  of  Algebra 
(Art.  235).     In  III.  of  Art.  153,  we  found  one  of  the  values 


of  (cos  ^  +  V  —  1  sin  $) ' ;  we  shall  now  find  an  expression 


which  will  give  all  the  q  values  of  (cos  6  +  V—  1  sin  0)''. 

Now  both  cos  6  and  sin  6  remain  unchanged  when  6  is 
increased  by  any  multiple  of  27r;  that  is,  the  expression 
cos^  +  V— Isin^  is  unaltered  if  for  6  we  put  (d-\-2rTr), 
where  r  is  an  integer  (Art.  36). 


(cos^  +  V^^sin^)' 


.1*    ''■!> 


=  [cos(e  +  2r7r)  +  V-lsin(^4-2?>)]'     '     '        '" 

p{e  +  2r7r)   ,     / — ^    .    p{d-h2r7r)   .,    ,  V-qwVx 
=  cos-^--^ ^+V~lsm^-^^ (Art.lo3)(l) 

The  second  member  of  (1)  has  q  different  values,  and  no 
more;  these  q  values  are  found  by  putting  r=0,  1,  2,  •••  7  — 1, 
successively,  by  which  we  obtain  the  following  series  of 

angles. 

?)(^+2rjr)  pd 

When  r  =  0,       cos — - 


=  cos 


<t 


t( 


r  =  l, 

r  =  2. 


« 


« 


=  cos 


=  COS 


p{e-\-2  tt) 


p{e  +  4.n) 


etc  etc. 


232 


PLANE  TRIGONOMETRY. 


When  r=g— 1,  cos^-^-      —  -  —  cos^-^^ ^^ '-^ 

p(^  +  2grr-2    ) 
=  cos ^ 

All  these  q  values  are  different. 

p{e  +  2rir)            p{e-^2q-rr) 
When  r  =  q,  cos =  cos 


'  fpB     ^    \         pe 

=  cos  (  y  +  2p7r  1  =  COS  y, 


the  same  value  as  when  r  =  0. 


When  r  =  g  +  1,  cos  — ^ — =  cos 

=  C08?L£±M, 


iHi! 


the  same  as  when  r  =  1, 


etc., 


from  which  it  appears  that  there  are  q  and  only  q  different 
values  of  cos^-^^ ~,  since  the  same  values  afterwards 

q 

recur  in  the  same  order. 

«..,,„        .   p{e-^2rT) 
Similarly  for  sm • 

Therefore  the  expression 
p{e-^2  rn) 


COS 


f  V—  1  sin 


p(e  +  2r^) 


gives  all  the  q  values  of  (cos5  + V—  1  sin^)'  and  no  more. 
And  this  agrees  with  the  Theory  of  Equations  that  there 
must  be  q  values  of  x,  and  no  more,  which  satisfy  the  equa- 
tion 3^  =  0,  where  c  is  either  real  or  of  the  form  a+6  V— 1. 


APPLICATIONS  OF  DE  MOIVRE'S  THEOREM.    233 


pe 


APPLICATIONS  OF  DE  MOIVRE'S  THEOREM. 

155.  To  develop  coand  and  BinnO  in  Powers  of  sin 6  and 
cos  6. 


We  shall  generally  in  this  chapter  write  i  for  V—  1  in 
accordance  with  the  usual  notation. 

By  De  Moivre's  Theorem  (Art.  153)  we  have 

cos?t^  + ^  sin?j^  =  (cos^  +  isin0)"  .     ,     .     .     .     (1) 

Let  w  be  a  positive  integer.  Expand  the  second  member 
of  (1)  by  the  binomial  theorem,  remembering  that  i^=— 1, 
i?  =  —  i,  and  that  i'^  =  -fl  (Algebra,  Art.  219).  Equate 
the  real  and  imaginary  parts  of  the  two  members.     Thus, 

cos  nd  =  cos"  6  -  ^  ^^^  ~  ^)  cos"-2  $  sin^  $ 

n(n-  1)  (n  -  2)  (n  -  3)  ^^^,._,^  ^.^^,^_  ^^  ^o) 

sin  nd  =  n  cos"  "'^  sin  6  -  ^^0^  ~  ^)  (^^  ~  2)  cos"-'^^  sin'd 

n(n-l)  {n-2)  (h-3)  (.i  -  4)  eos.-5^,i„.^_etc.  (3) 

The  last  terms  in  the  series  for  cos  nO  and  for  sin  n$  will 
be  different  according  as  n  is  even  or  odd. 

The  last  term  in  the  expansion  of  (cos  d  +  tsin^)"  is 
/"sin"^;  and  the  last  term  but  one  is  ?u"  ^cos  tf  sin"~'d. 
Therefore : 

When  n  is  even,  the  last   term  of  cos?i^  is  t''sin"^  or 

n 

(  —  1)^  sin"^,  and  the  last  term  of  sin  nd  is  ni"~^  cos  0  sin"~'tf 

n-2 

orn(— 1)*   cos^sin""^^. 

When  n  is  odd,  the  last  term  of  cos  n6  is  wi"~^  cos  6  sin"~'^ 
..-1 

or  w(— 1)  '■'   cos^sin""'^,  and  the  last  term  of  sinnO  is 

n-l 

t'-'sin"^or  (—1)  *  sin"^. 


234 


PLANE  TRIGONOMETRY. 


EXAMPLES. 

Prove  the  following  statements : 

1.  sin 40  =  4: cos^ ^  sin  ^  —  4 cos 6  sin^ $. 

2.  cos 4 ^  =  cos* 6-6 cos^ 6 sin^ $  +  sin* 0. 


iiiiii> 


156.  To  develop  sin  6  and  cos  6  in  Series  of  Powers  of  6. 

Put  nd  =  a  in  (2)  and  (3)  of  Art.  155;  and  let  n  be  in- 
creased without  limit  while  «  remains  unchanged.     Then 

since  6  =  -,  0  must  diminish  without  limit.     Therefore  the 
n 

above  formulae  may  be  written 

cos  a  =  COS"  e  -  '^^=^  cos"-^  6  f'^^" 

\2_  \   d   J 

4.«(«-^)('^-2^)(«-3^)cos"-*g(g^y-...      (1) 

and  sin  «  =  «cos"~*^( ] 

\   0   J 


_,(^_^)^(^_2g^^^^,._3    /sin^3_^ ...   . 


(2) 


If  w  =  QO,  then  ^  =  0,  and  the  limit  of  cos  0  and  its  powers 

is  1;  also  the  limit  of  (     -— )  and  its  powers  is  1.     Hence 
(1)  and  (2)  become      ^        ^ 


a'  .  a* 


« 


cos«  =  l 1- , , — !-••• (3) 


2    '14      [6 


Sin  a  =  «  — , — •  •  • 


.']  ■  [5_ 


•         ••••••#•  \     / 


Sch.   In  the  series  for  sin  a  and  cos  «,  just  found,  a  is  t?ie 
circular  measure  of  the  angle  considered. 


CONVERGENCE  OF  THE  SERIES. 


235 


Cor.  1.  If  a  be  an  angle  so  small  that  a^  and  higher 
powers  of  a  may  be  neglected  when  compared  with  unity, 
(3)  becomes  cos  a  =  1,  and  (4),  sin  «  =  «. 

If  «^,  u^  be  retained,  but  higlier  powers  of  a  be  neglected, 
(3)  and  (4)  give 

sin  a  =  a  —  — ;  cos  a  =  1  —  —    (Compare  Art.  134) 
Cor.  2.   By  dividing  (3)  by  (4),  we  obtain 


^3      2  a* 

3      3  •  o      3  •  3  •  o  •  < 


17  «^       ,      , 

—_  +  etc. 


(5) 


157.  Convergence  of  the  Series.  —  The  series  (3)  and  (4) 
of  Art.  156  may  be  proved  to  be  convergent,  as  follows : 

The  numerical  value  of  the  ratio  of  the  successive  pairs 
of  consecutive  terms  in  the  series  for  sin  a  are 


a*        «'         «* 


«' 


2-3    4-5    0-7    8-9 


etc. 


Hence  the  ratio  of  the  (n  -f  l)th  term  to  the  nth  term  is 
-;  and  whatever  be  the  value  of  a,  we  can  take 


2w(2«  +  l) 

n  so  large  that  for  such  value  of  n  and  all  greater  values, 

this     action  can  be  made  less  than  any  assignable  quantity ; 

hence  the  series  is  convergent. 

Similarly,  it  may  l)e  shown  that  the  series  for  cos  a  is 

always  convergent.  -  -  " 

158.  Expansion  of  cos"^  in  Terms  of  Cosines  of  Multi- 
ples of  6,  when  n  is  a  Positive  Integer. 

Let     .r  =  cos^  +  t  sin^; 

1  1 


then 


X 


cos  $  -\-i  sin  6 


=  cos  6  ~  i  sin  $. 


.:  a;  +  -  =  2 cos 6 :  and  x  — =2i sin 6 

X 


X 


(1) 


236 


PLANE  TRIGONOMETRY. 


i^«ii 


liiiii 


if 


Also  a;"=(cosd+isin^)"=cosn^4-tsmwd  (Art.  153)  (2) 
and       —  =  (cos  6  —  i  sin  6) "  =  cos  nd  —  i  sin  nB  .     .     .     (3) 

1  1 

.'.  2 cos n^  =  a;"H — .and  2isin?i0  =  a5" .     (4) 

Hence  (2cos^)"  =  (a;  +  .'K-i)",  by  (1), 
=  a;"  +  nx"--  +  '*^^^^  —  ^)  a?'-*  -f-  etc.  +  «a;  <"-2)  +  aj-" 


n 


(n-1) 


==  2  cos  wtf  +  n  2  cos  ( ji  -  2)  e  +  "v,     "'^  2  cos  (n  -  4)  ^  +  etc. 

If 

.-.  2"-^  cos"  d  =  cos  91^  +  w  cos  (w  —  2)  ^ 

+ '^-^^^^-^^  cos  (?i  -  4)  ^  +  etc.        (5) 

Note. — In  the  expansion  of  (a; +  «"■')"  there  are  n  +  1  terms;  thus  when  n  is 
even  there  is  a  middle  term,  the  |  -  +  l)th,  which  is  independent  of  9,  and  which  is 

n(n-l)...(n-4n  +  l)  ^^    ^2       / 

lln  l-i" 

Hence  when  n  is  et'en  the  last  term  in  the  expansion  of  2"^'  cos**  6  is 

Hi 

When  n  is  odd  the  last  term  in  the  expansion  of  2"'^  cos"  9  is 

159.  Expansion  of  sin" '}  in  Terms  of  Cosines  of  Multiples 
of  6,  when  n  is  an  Even  Positive  Integer. 

(2  i  sin  ^)»  =  /'a;  -  ^V'by  (1)  of  Art.  158 

=  a;"  -  n^-''  +  -^^""^^  «"-♦  +  .- 

4.  ^_(^  1)  a.-("-4)  _  ,,x'-('-'»>  +  x-^ 


II' V. 


(5) 


EXPANSION  OF  SIN"  6. 


237 


=e-9-"e-^)-'^e*^^; 


(-irn(n-l)...(:-;  +  l 


+  •••  + 


n 


[i± 


..  2"-'(— l)'^sin"0 

=  cos  w^  -  n  cos  (h  -  2)  ^  +  "^''~^^  cos  (n  -  4)  J 

If 


(-irn(^-i)-  :^+i 


+ 


ri 


^Li± 


160.  Expansion  of  sin"  6  in  Terms  of  Sines  of  Multiples  of 
0,  when  n  is  an  Odd  Positive  Integer. 


(2  i  sin  0)"  =  (^a;  -  -Yby  (1)  of  Art. 


158 


z=x"  —  nx' 


n-2   I   n  (n  -  1)  ^,._4 

[2 


X' 


n-4 


n-1 


li(n-i)         V    ^; 

.-.   (2isin0)" 

=  2isinri0  — n2isin(n  — 2)6 

n(n-l)2,-sin(n-4)6 

^       [2 

.1-1 
(-1) '^  n(w-l)---K^^±^9.-«infl    [(4)  of  Art.  158] 


1S¥ 


238 


PLANE  TRIGONOMETRY. 


Whence  dividing  by  2t,  we  have 

w-l 

2-'(_l)  2  sin"^ 

=  sin  110  -  71  sin  (n  —  2)tf  +  ^^-f-^  sin  (71  _  4)^  +  .- 

(-1)^.(n^)...^(»,  +  .S)^.^  ^^ 

l''(H-J) 


ll 


» 


-^ 


EXAMPLES. 
Prove  that 

1.  128  co3''^=cos  Se+Hcos(5e+'2S  cos  4^+56  cos  20+35. 

2.  64  cos^0  =  cos  70  -f  7  cos  50  +  21  cos  30  +  35 cos  0. 

161.  Exponential  Values  of  Sine  and  Cosine. 

Since     e'=l  +  a;  +  ^  +  ,^'  +  ^'+-     .     .    .     (Art.  129) 

Lf     L?     Li 

0' 


•   p«  —  1  _  !L  4.  ^ 

••  ^    "^      [2  +  " 

=  cos  0  +  i  sin  0 


and 


(Art.  156) 

=  cos  6  —  i  sin  0. 
.-.  2  cos  0  =  e*"  +  e-  *«,  and  2  i  sin  0  =  e»«  -  e-»«     .     .     (1 ) 


COS0  = 


e*'*  +  e 


-i0 


-,  and  sin  0  = 


otS 


ie 


2t 


•         • 


(2) 


which  are  called  the  exponential  values*  of  the  cosine  and 
sine. 

Cor.   From    these    exponential   values   we    may  deduce 
similar  values  for  the  other  trigonometric  functions.    Thus, 

tan0  =  -i— -^-- (3) 

*  Called  also  Euler's  equations,  after  Euler,  tbeir  discoverer. 


6  + 


^rt.  129) 


\rt.  156) 


.  .  (1) 
.  .  (2) 
Dsine  and 


y   deduce 
IS.    Thus, 

.    .     (3) 

jr. 


f 


GREC,  on  Y '  8   SERIES. 


239 


Sell.  These  results  may  i)e  applied  to  prove  any  general 
fornnda  in  elementary  Trigononu^try,  and  are  of  great  im- 
portance in  the  Higher  Mathematics. 


BXAAuPLBS. 


1 .    Prove 
We  have 


sin2g 
l  +  cos2^ 

2tsin2g 

2  +  2cos2tf~2  +  e2«  +  ^ 


=  tan  6. 


o2i» 


ii» 


2i» 


by  (1) 


pi9 g-  %9 

Prove  the  following,  by  the  exponential  values  of  the 
sine  and  cosine. 

2.  cos  2  rt  =  cos^ «  —  sin'' a. 

3.  sin^  =  — sin(  — ^). 

4.  cos3^  =  4cos''e-3costf. 

Rem.— It  we  omit  the i  from  the  exponential  values  of  the  eine,  coBine,  and  Un- 
gent  01  0,  the  reaulta  are  called  respectively  the  hyperbolic  sine,  cosine,  and  tangent 
of  9,  and  are  written  sinh  9,  cosh  9,  and  tanh  9,  respectively.    Thus  we  have 
■inh  0  =  —  f  sin  ifl,  cosh  9  =  cos  i9,  tanh  0  =  —  i  tan  i9. 

Hyperbolic  functions  are  so  called,  because  they  have  geometric  relations  with 
the  equilateral  hyperbola  analogous  to  those  between  the  circular  functions  and  the 
circle.  A  consideration  of  hyperbolic  functions  is  clearly  beyond  the  limits  of  this 
treatise. 

For  an  excellent  discussion  of  such  functions,  the  student  is  referred  to  Buch 
works  as  Casey's  Trigonometry,  Hobson's  Trigonometry,  Lock's  Higher  Trigonom- 
etry, etc. 

162.  Gregory's  Series.  —  To  expand  6  in  jiowers  of  tan  $ 

where  6  lies  between  —  -  and  4-  -• 

2  2 


By  (3)  of  Art.  161,  we  have 


i  tan  0  = 


-»■« 


l+ttan^^  26'" 
1-iidLue     2e-»« 


=  e2»o. 


240 


PLANE  TRIGONOMETRY. 


i 


.-.  log  e2»<'=  log(l  +  i  tan  6)  -  log(l  -  i  tan  0). 

. :  2  W  =  2  <■  (tan  tf  -  ^  tan'^  6  +  ^  tan'  0  -  etc. )    (Art.  130  ) 

.-.  ^=tan^- ^tan^0  +  |tan*^-etc (1) 

which  is  Gregory^s  Series. 

This  series  is  convergent  if  tan  ^  =  or  <  1,  i.e.,  if  6  lies 


between 


v 


^  and  -,  or  between  ^tt  and  Itt. 
4         4 


Sch.   This  series  may  also  be  obtained  by  reverting  (5) 
in  Cor.  2,  Art.  15(1 


Cor.  1.    If  tan  6  =  x,  we  have  from  (1) 


x^  .  x* 


tan~^a;=  x 1 etc. 

3      6 

Cor.  2.   If  ^  =  -,  we  have  from  (1) 
4 


(2) 


IT 


^  =  l-i  +  i-|  +  etc. 


(3) 


a  series  which  is  very  slowly  convergent,  so  that  a  large 
number  of  terms  would  have  to  be  taken  to  calculate  tt  to  a 
close  approximation.  We  shall  therefore  show  how  series, 
which  are  more  rapidly  convergent,  may  be  obtained  from 
Gregory's  series. 


163.  Enler's  Series. 

.     .     (by  Ex.  2,  Art.  60) 

Puttf  =  tan-ii.    .-.  tan  (9  =  ,^,  which  in  (1)  of  Art.  162 


tan-'  -  +  tan-'  ^  =  - 
2  3     4 


gives 

tan-'  -  =  - —  +  — +  etc.  . 

2     2     3.23^6.2*     7-2' 


•         •         • 


(1) 


Put  $  —  tan 


'  -    .'.  tan  ^=-,  and  (1)  becomes 
3  3  ^  ^ 


tan  *  -  = H ;  +  etc. 

3     3     3.3^     6.3'     7.3^ 


•        •         •         •         • 


(2) 


MACHINES  SERIES. 


211 


(2) 


.      (2) 


Adding  (1)  and  (2)  we  have 

4     V2     3-23     5-2*         J     \3     3.3^     5.3*         J    ^  ^ 

a  series  which  converges  much  more  rapidly  than  (3)  of 
Art.  162. 

164.  Machines  Series. 

Since  2tan-^^=tan -^--i-  (by  Ex.  3,  Art.  60)=tan-»-|, 
5  1  —  Alt  i^ 


•^3 


.-.  4  tan-'  -  =  2  tan"'  4  =  tan"'  — i—  =  tan"'  ^ 


-.1 
5 


12 


1    _     2.5 


Also,    tan-'  ^  -  tan"'  1  =  tan"'  ^    }  =  tan 


119 
...  4tan-'i-T=tan- 


5     4 


=  4  tan 


239 


119 
239 


tan" 


239 


.7r_./l__l ^1__      \ 

"I        \5     3.53     5-5*     "V 


-M    _ 


+ 


-i Y 

;239)*      y 


V239     3.(239)»     5(239)* 
In  this  way  it  is  found  that  ir  =  3.141592653589793  ••.. 

Cor.   Since  tan-'  1  +  tan"'  -1-  =  tan"'  ^^-t^ 

NoTB.  —  The  BcricB  for  tan-'  ^   and  tan"'  '-  are  much  more  convenient  for  pur- 

70  99 

poses  of  numerical  calculation  than  the  series  for  tan~'  — -• 

239 

Example.  —  Find  the  numerical  value  of  tp  to  6  figures  by 
Machin's  series. 


amimm^mgeom 


I  - 

If. 


rt 


242 


PLANE  TRIGONOMETRY. 


165.  Oiven  sin  9  =  x  sin  {6  +  a);  expand  d  in  a  Series  of 
Ascending  Powers  of  x. 

]  .    .     (Alt.  161) 


We  have 

gi«_ 

e-»«  = 

x[e^^+ 

ia 

e- 

-i9 

-ia 

.-.  e2»» 

-1  = 

X  [e2»» . 

gia 

— 

e- 

*•] 

« 
•  • 

Q2i9- 

1  —  xe- 

-ia 

1  —  ice*'" 
.-.  2'i$  =  log(l  —  xe-*")  —  log(l  —  xe^) 

=  a;(e»«»  —  e-»")  +  ^(e2»a_g-2ia)  _j_  ^(^^ia_e-zia^  ... 
^  ^         (Art.  130) 

.-.  ^  =  a;sina4-f  sin2«  +  ^sin3a+---  (Art.  161) (1) 

Example.   If  a  =  tt  —  2^,  then  a;  =1.     .*.  (1)  becomes 
^  =  sin  2^  -  ^  sin40  +  \  sin 6^  -  \  sin 8^  +  •••. 

166.  Oiven  tan  a;  =  n  tan  d ;  expand  x  in  Powers  of  n. 

g»9 g  -  i9 


gia?  _  g-»x 
gir  _|_  g-w: 

/>2ta; . 


=  71- 


plix 


+  1 


=  71 


gi«  -I-  g-W 
g2i«  __  1 


(Art.  161) 


,'iie 


+  1 


.  g2te  =  i^!!!±l)±ii(f'*!zill 

^e2ifl_|.l)_„(g2»9_l) 
_  (1  +  M)e2f9  -f  1  -  n 

(l-7i)e2»»  +  l4-n 
^g2«_|_i  \^  1  +  nJ 

.-.  2ta;  =  2  ttf  +  log(l  +  me-  2«)  -  log(l  +  me'^i«) 

=  2td  -  m(e2»»  -  e-2»«)  -|-  !^  (e4W  _  g-4W) . 


m" 


a;  =  ^  — msin2d4-  — sin4tf 


(Art.  161) 


0  \\\ 


RESOLUTION  INTO  FACTORS. 


243 


RESOLUTION  OF  EXPRESSIONS  INTO  FACTORS. 

167.  Resolve  x"  —  1  into  Factors. 

Since  cos  2  ?*7r  ±  V  —  1  sin  2  rTr  =  1, 

where  r  is  any  integer,  and  of  =  1, 


.•.  ic"  =  cos  2?*7r  ±  V—  1  sin  2r7r. 

1^ 

.-.  a;  =  (cos2r7r  ±V-lsin2r7r)" 


2r 


2rTr' 


=  cos ±  V—  1  sin 

n  n 


(Art.  163)  (1) 


(1)  When  n  is  even.   If  r  =  0,  we  obtain  from  (1)  a  real 


root  1 J  if  r  =  -,  we  obtain  a  real  root  —  1,  and  the  two  cor- 
responding  factors  are  x—l  and  x  +  1.    If  we  put 

r  =  l,2,3...^-l, 

in  succession  in  (1),  we  obtain  ri  — 2  additional  roots,  since 
each  value  of  r  gives  two  roots. 

The  product  of  the  two  factors,  which  are 


and 


(X  —  cos  —  —  V—  1  sin 
n 

(x  —  cos  —  +  V^^  sin 
V  n 


2r7r\ 
n  ) 


2r^■^ 


n 


( 


■=.{  05  — cos- 


n  J 


4-  sin'' 


n 


=  ar  — 2a;co8 hi 

n 


(2) 


which  is  a  real  qiiadratic  factor. 

.-.  a;"-l=  (o;2-l)raj*-2a; cos ^  -j-lYa;'- 2xcos ^  +  iV- 

../ar'_2a;cos*^7r+lY.x'-2a5C08^7r-|-lY..  (3) 

*  Thia  eipreaaloD  give*  the  n  nth  root*  of  unity. 


Ff 

m 

!  i, 

1* 

244 


PLANE  TRIGONOMETRY. 


(2)    TF/ien  )i  is  odd.     The  only  real  root  is  1,  found  by 
putting  »'  =  0  in  (1);  the  other  71  —  1  roots  are  found  by 

71—1 

putting  r  =  1,  2,  3, —  in  (1)  or  (2)  in  succession. 

.'.  af— l  =  (a5— l)f  ar'— 2a;cos— +  1  jf  ar^— 2a;cos  — -f- 1  j-'- 
^-far'— 2a;cos  TT-fl  jf  ar*— 2a;cos tt+I  ]•••  (4) 


li 


Ml 


i     M 


11 


lit) 


168.  Resolve  a;"  4- 1  into  Factors. 


Since     cos  (2r  +  l)7r  ±  V—  1  sin  (2r  +  l)7r  =  —  1, 
where  r  is  any  integer,  and  a;"  =  —  1, 


.-.  af  =  cos  (2r  +  l)7r±V— 1  sin  (2r +  !)«•. 


.-.  X  =  [cos  (2r  4-  l)7r  ±  V—  1  sin  (2r  -j-  l)7r]" 

2r+l           .— ^    .    2r+l 
siCOS 7r±V— ISin TT  .      .      . 


(1) 


n  n 

which  is  a  root  of  the  equation  a;"  =  —  1_;  i.e.^  —  1  is  a  root. 

(1)    When  n  is  even.     There  is  no  real  root;  the  n  roots 
are  all  imaginary,  and  are  found  by  putting 

r=:0,l,2,...^-l, 


successively,  in  (1), 

The  product  of  the  two  factors. 


(: 


X  —  cos  - 


2r  +  l 


n 


—  V—  1  sin 


'-■^') 


and 


I  n        '  **      / 

=  a^-2a;co8?ili^7r4-l (2) 

n 

which  is  a  real  quadratic  factor. 


RESOLUTION  INTO  FACTORS. 


245 


(4) 


.-.  cc"  +  l  =  /V-2a;cos^  +  lYar-2a;coSy^  +  lY.. 

../ar'  -  2a; cos '-^^tt  +  iVar^ -  2a; cos  ^^ tt  +  lY-   (3) 

(2)    When  n  is  odd.     The  only  real   root   is   —  1 ;   the 

?t  —  3 
other  n  —  1  roots  are  found  by  putting  r  =  0,  1,  2,  •  •  • 

in  (1),  in  succession. 

.-.  a;»+l=  (a;  +  l/a^  -  2a; cos  ^  +  lYa-^  _  2a; cos  ^  +  iV- 


•  ••[ar^  — 2a; 


cos^?— li7r  +  lY^-^-2a;cos--^7r  +  l']   (4) 


71 


n 


EXAMPLES. 

1.  Find  the  roots  of  the  equation  ar*  —  1  =  0. 

Aus.  1,  cos  \  (2  ?'7r)  +  i  sin  \  (2  rir),  where  r  =  1,  2,  3,  4. 

2.  Find  the  quadratic  factors  of  a^  —  1. 

Ans.  {x"  -  1)  (ar^ -  ^2 a;  +  1)  (^  +  1)  (^f^  +  v'2 a;  +  1). 

3.  Find  the  roots  of  the  equation  a;^  +  1  =  0,  and  write 
down  the  quadratic  factors  of  x*  -\- 1. 

■Ans.  ±  —  ±  V^^-^;  (a^  -  a;V2  +  1)  (x-  +  a;V2  +1). 


V2 


V2' 


169.  Resolve  a;^  —  2  x"  cos  ^  + 1  into  Factors. 

Let    a;^-2a;"cos^  +  l  =  0. 
.  •.  x^"  -  2  X"  cos  6  +  cos^  e  =  —  sin^  0. 

.'.  a;"  —  cos 6=  ±  V—  1  sin  ^  =  ±  «  sin $. 


.:  X  =  (cos  B  ±i  sin  ^)"  =  cos 


2rw  +  0 


±  I  sm 


2rw  +  e 


(1) 


7t  n 

since  cos  d  is  unaltered  if  for  0  we  put  6  -\- 2  rir.  If  we  put 

r  =  0,  1,  2,  •••??  — 1,  successively  in   (I),  we  find  2n  differ- 
ent roots,  since  each  value  of  r  gives  two  roots. 


p 


lill'l 


ft 


!|  III 


!f   ift' 


^1 

lii 


:r    .ill 


lifii 


246 


PLANE  TBWONOMETRY. 


The  product  of  the  two  factors  in  (1) 


cos  — 


X  I  05  —  cos ' — 

\  n 


2rn  +  d      ..2r7r-^b 


—  I  sin 


+ 1  sin 


2rir  +  e 


^) 


=  ar2-2a;cos^':^^:+-^  +  l 

n 

.*.  05**  —  2  af  cos  ^  +  1 

=  /^a^-2a;cos^  +  iyg'-2a;cos^^  +  ^  +  lY.. 

r          ^            (2»i-4)7r  +  d        \ 
...far'-2a;cos^^ J  +ll— 


(2) 


/»«- 


2ajcos 


(2n-2)7r  +  d 


n 


+  1 


!•••  .    .     .     . 


(3) 


a; 


Cor.   Change  x  into  -  in  (3)  and  clear  of  fractions,  and 


a 


we  get        a;**  —  2  a''af*  cos  tf  +  a**"  =  [  ar*  —  2  aa;  cos --\-aA'-' 
../ar' -  2  oa;  cos  ^^!^±^  +  a'Va;' -  2  aa;  cos  i^^^JtJ  ^_  a^y 

•  ••  to  w  factors (4) 

EXAMPLES. 

Find  the  quadratic  factors  of  the  following : 

1.  a;*- 2a;* cos 60°  4- 1  =  0. 

Ans.  {it'  -  2x  cos  15°  +  IXx**  -  2  a;  cos  105°  +  1) 

X  (a«  -  2  a?  cos  195°  +  1)  (ar*  -  2  a;  cos  286°  +  1)  =  0. 

2.  a;»°-2a'»cosl0°  +  l=0. 

Ans.  («« -  2 a;  cos  2°  4- 1)  (x'  -2x  cos  74°  +  1) 

X  (ar*  -  2  a;  cos  146°  +  1)  («2  _  2  a;  cos  218°  4- 1) 
(ar' -  2  X' COS  2yo" -f  1)  =  0. 


DE  MOIVRE'S  PROPERTY  OF  THE  CIRCLE.      247 


(2) 


170.  De  Moivre's  Property  of  the  Circle 

centre  of  a  circle,  1*  any  point  in  its 
plane.  Divide  the  circumference  into 
n  equal  parts  BC,  CI),  DE, -..,  begin- 
ning at  any  point  B;  and  join  O  and 
P  with  the  points  of  division  B,  C, 
D,  •••.     Let  FOB  =  0 ;  then  will 


OB'"  -  2  0  B" .  OP"  cos  ne + OP 


:2n 


= PB'  .  PC  .  PD  •  ■ .  to  ?i  terms. 
For,  put  OB  =  a,  OP  =  x,  and  $  =  -;  then 
PB'  =  OP  +  OB'' -  20P  •  OB  cos  $ 


=  ar*  +  a^  —  2  ax  cos  - 

n 


(1) 


;2      n/-wT^    r\/~i a-[-2ir 


PC'  =  oF  +  OC  -  20P  ■  OC  cos 

=  x^  ^a'-2ax  cos  "-+15.;  and 


so  on 


(2) 


'ax 


i^  +  a" 


Multiplying  (1),  (2),  (3),...  together,  we  have 
PB'  ■  PC'  •  PI? .••  to  n  terms 

^zfx'  -  2ax  cos  "  +  aNx"  -  2 

=  a:2«  _  2a''a;»  cos  «  +  a^"    .     [by  (4)  of  Art.  169] 

=  OP«»-20P"-OB"cos7i^  +  OB'"     .    .     .     (3) 
which  proves  the  proposition. 


ip' 


,      !■' 


M'i 


m ! 


"J   !'l 

I!! 
'I?  i'! 


m'i 


I 

I'll 


ft  ;i! 


248 


PLANE  TRIGONOMETRY. 


171.  Cote's  Properties  of  the  Circle.  —  These  are  particu- 
lar cases  of  De  Moivre's  property  of 
the  circle. 

(1)  Let  OP,  produced  if  necessary, 
meet  the  circle  at  A,  aud  let 

AB  =  BC  =  CD,  etc.,  =  -'!'; 

n 

then  n$  is  a  multiple  of  2  v.     Hence 

we  have  from  (3)  of  Art.  170,  after  taking  the  square  root 

of  both  members, 

OB"  -  OP"  =  PB.  PC.  PD...  ton  factors.     .     .     I. 

(2)  Let  the  arcs  AB,  BC,  "-be  bisected  in  the   points 
a,  6,  ••• ;  then  we  have,  by  (1), 

OB'"  -  OP*"  =  Pa  .  PB  .  P6  .  PC  .  Pc  ...  to  2?i  factors. 


Hence,  by  division. 


IL 


OB"  4-  OP"  =  Pa  .  P6  .  Pc . . .  to  ?i  factors  .     . 

Cor.   If  the  arcs  AB,  BC, ...  be  trisected  in  the  points 
a,,  a2,  6i,  62,  ••.,  then  we  have 

OB'"+OB" .  0P''+0P2''=Pax . Paj,  •  Vb^ . P62 ...  to  2»i  factors. 

172.  Resolve  sin  6  into  Factors. 

(1)  Put  a;=  1 ;  then  we  get  from  (3)  of  Art.  1G9 

2(l-cos^)=2"/'l-cos^/'l-cos^±^Yl-cos^-^±^Y.. 


'.'(1 
Put  tf  =  2?i<^  in  (1),  and  let  2n«  =  7r. 


n     j\ 


(1) 


1  —  co8^  =  1  —  cos2w<^  =  2sin''n<^; 

then  extracting  the  square  root,  we  have 

Bin?}<^  =  2"-'sin<^  •  sin  («^  4- 2rt)sin  (<^ -f  4«)  x 

...  X  sin  (</>-{- 2n«  —  2«) (2) 


I    ^ 


RESOLUTION  INTO  FACTORS. 


249 


•ticu- 


vB 


U 


root 


I. 


.     (2) 


But  sin  (tf)  +  2n«  — 2«)  =  sin(<^  +  7r  — 2«)  =  sin(2rt  — <;^), 
sin  {<f} -\- 2 na  —  4 a)  =  sin(4«  —  <f>),  and  so  on. 

« 

Hence,  when  n  is  odd,  multiplying  together  the  second 
factor  and  the  last,  the  third  and  the  last  but  one,  and  so 
on,  we  have 

sinnt^ 

=2"~'sin<^sin(2rt+<^)sin(2«— <^)sin(4«-f<^)sin(4«— «^) 
•  ••  X  sin[(w  — 1)«  +  <^]  sin  [{n  —  l)u  —  <^]. 

But  sin  (2«-|-<^)  sin  (2  a— <^.)  =  sin^2«  — sin^<^,  and  so  on. 

.*.  sinn^=2''~'sin  <^(sin''2«  —  sin*«^)(siu^4«  —  sin-<^)  X  ••• 
..•  X  [sin-(n  —  1)«  — sin'-'i/)]     ....     (3) 

Divide  both  members  of  (3)  by  sin  <^,  and  then  diminish 
<^  indefinitely.  Since  the  limit  of  sin  n<f>  -4-  sin  <t>  is  n,  we 
get 

n  =  2""*sin^2rt  sin^4«  sin'^Ga  x  •••  X  sin'^(w  —  1)«     (4) 

Divide  (3)  by  (4) ;  thus 

sm  7n^  =  n  sm  <t>{l  —  ^-„-„^      1  -   .   .,/      X  •  •  •     (5) 

Put  n<fi  =  6,  and  let  n  be  increased  while  <^  is  diminished 
without  limit,  6  remaining  unchanged;  then  since  2nu  =  ir, 
tl^  limit  of 


8iu'<^ 
sin*  2  a 


sm' 


,$ 


^ 


sin 


0 


SUV 


n 
-  =  -,  X 

n 


n 


=  ^  (Art.  133) 


sm 


0 


jjTT  IT 

n 


and  the  limit  of  n  sin  <b  =  that  of  n  sin  -  =  6:  and  so  on. 

n 

Hence  (5)  becomes 

Note.  —The  aame  reeult  will  be  obtained  if  we  auppoae  n  even. 


250 


PLANE  THIGONOMETHY. 


i 


I 

i 
11 

II 

i 

1  !■! 

1 

v' 
f 

j    ;: 

\ 

,    1 

•ifi; 

\ 

t 

.  _--. 

:     '; 

■( 

% 

!  11 

'  1      ■, 

} 
1 

'  1= 

;     i 

5     -\ 

!    !!! 

■1 

^       1 

!        i 

I 

i 

1 

•  Rem.  —  When  fl>0  anJ  o,  •Intf  ia  +,  und  every  fitctor  in  the  •ecoi  nber 

of  (0)  ii  poitltive;  when  0>ir  and  <2ir,  ■in0  \*  -,  und  only  the  leconU  '•  U 

negative;  wlieu  9>2irand  <3;r,  botli  menibem  are  positive,  Hinee  only  tht  >ad 
and  third  factors  are  negative;  and  so  on.  Hence  the  +  sign  was  talten  in  lAiract- 
ing  the  square  root  of  (1),  , 


comes 


Cor.    Let  6  =  "^,  then  sin  -  =  1,  and  -  =  ^.     Hence  (6) 


be- 


^=i^ 


^    /  \  ^     *  ^  J 

TT  1-3  a.  5  5-7 

;  -_  •  I  I  a     ■    ■■  -  .  • •  •  • 

2     2^      4^'      & 
22       42        62        82 


2     1.3  3.5  5.7  7.9 
which  is  Wallis^s  expression  for  tt. 

173.   Resolve  cos^  into  Factors. —  In   (2)  of  Art.  172, 

change  </>  into  </>  +  «,  then  n<\>  becomes  n<\>-\-  nu,  i.e.,  n<fi  -f-  -• 
Hence  (2)  becomes 

cos  n<ft  =  2"~'  sin  (<f>  +  «)  sin (<^  4-  3  «)  sin  (<^  +  5  «) 

X  •••  sin[</>  +  (2?i  — l)rt] (1) 

But  sin(<^  -f  2?i«  —  a)  =  sin(<^  -j-  tt  —  «)  =  sin(«  —  <{>), 
sin(<^  +  2nH  —  3«)  =  sin(3«  —  <^),  and  so  on. 

Hence  when  n  is  even  we  have  from  (1) 
cosn<^=2"'*  sin(rt-f  <^)  sin(«  —  6)  sin(3«  +  </>)sin(3rt  —  ^) 
X  •••  X  siu[(n  —  1)«  -f  <^]sin[(?i  —  !)«  —  </>] 

=  2"-'(sin«  rt  -  s'm^)  (sin*3«  -  sin^) 

X  •••  X[sin''(n-l)«-sin*(^] (2) 

Therefore,  putting  n<t>  =  6,  as  in  Art.  172,  we  obtain 


cos 


H'-'-S)H^ttQ 


(3) 


NoTK.  —  For  an  alternative  proof  of  the  propositions  of  Arts.  172  and  173,  see 
Locli's  Higher  Trigonometry,  pp.  92-95. 


SUMMATION   OF  SERIES. 


251 


EXAMPLES. 


IT 

4n' 


1.    If  rt  =  ."-,  prove  that 


sin  rt  sin  5  a  sin  9  « •  •  •  sin  (4  n  —  li) «  =  2  "+*. 

2.   Show  that 

16cosdcos(72°-d)cos(72°4-e)cos(144°-d)cos(144°  +  ^) 

=  cos  5  $. 

SUMMATION    OF    TRIGONOMETRIC    SERIES. 

174.  Sam  the  Series 

sin  It  +  sin(rt  +  /3)  +  sin(«  +  2/3)  +  •••  +  si"  ['«  +  («  -  1)^]- 
We  have 

2  sin  «  sin  ^  j3  =  cos  fa  -  ^  -cos  fu  +  |  j  (Art.  45) 

2  sin  (rt  +  i3)  sin^  j3  =  cos  fa  +  ^-  cos  (rt  +  f  ^), 

2 sin  (rt  -1-2/3)  sin  ^  /?  =  cos  (rt  +  1)3)  -  cos  (rt  -\-  f  ^), 

etc.  =  etc. 
2sin[rt-f(H-l)i3]sini^ 

Therefore,  if  S„  denote  the  sum  of  n  terms,  we  have,  by 
addition, 

2S„  sin  i/3  =  cos  (rt  -  ^/3)  -  cosU  +  -^^P 

=  2  sinfrt  4-  -■^-  iSl  sin  i wjS  .     .^  (Art.  45) 


sinfrt-i-'—^iS'lsin^niS 


sini/3 


252 


PLANE  TRIGONOMETRY. 


''*^ii  tidiii; 


|i' 


t   Mi 

ill 


'h:;; 


175.  Sum  the  Series 

cos  «4-co8  (rt4-/3)  +  eo8  («  +  2/8)  H +cos[rt4-(n— 1)/3]. 

We  have  2  oos  « sin  ^/3  =  sin  («  +  ^/3)  —  sin  («  —  ^y3), 
2  cos  (rt  +  /3)  sin  ^/3  =  sin  (a  +  |/3)  -  sin  («  +  ^y3), 
etc.  =  etc. 
2cos[rt  +  (?i-l)^]sin^/3 

=  8m   a  +  — —     ^    _sin    «  +  — ^^  . 

Denoting  the  sura  of  n  terms  by  S„,  and  adding,  we  get 
2S„sini/3  =  sinr«  +  ^^Y^/3l-sin(«-i)3). 

cos  I  «  +  — ■"-  /sl  sin  ^  ?i/3 


.-.  S„ 


sin^yS 


Bern.  — The  sum  of  the  series  in  this  article  may  be  deduced  from  that  in  Art.  174 
by  putting  a  +  -  for  a.    The  sums  of  these  two  series  are  often  useful;*  and  the 

■tudent  is  advised  to  commit  them  to  memory. 


iir 


Cor.   If  we  put  /3  =  — ,  then  sin  ^nft  =  sin  tt  =  0.    Hence 
we  have  from  Arts.  174  and  175 


sin 


cos 


rt+sinr«+— '^j  +  sin^<+y^j4-...sin   fi+ -^^ — -^  =0. 

«  +  C08[  — ]  +  COS(  U-\ )  +  ••• 


+  oosfrt+ '"')  +  •••  cosr«4-^^^^-^'rl=0. 

Note.  — These  two  results  are  very  important,  and  the  student  sliould  carefully 
notice  them. 

176.  Supa  the  Series 

sin'"«4-sin'"(«4-/3)+3in'»(«+2/3)H l-sin- [«+(«- l)i3]. 

This  may  be  done  by  the  aid  of  Art.  159  or  Art.  160. 

*  See  Thompson's  Dynamo- Electric  Machinery.    3d  ed.,  pp.  345,  346. 


■ 


SUMMATION  OF  SERIES. 


258 


Thus,  if  m  is  even,  we  liave  from  Art.  159 

2m-i  gin- rt  =  (-1)2  [cos  m«-m  COS  (wi-2)«4-...]    .     (1) 
2*-^  sin"*  («  +  (3) 

m 

=  (— 1)" [cos m  («+/3)  — w cos  (w-2)  (rt4-/3)  +  •••]      (2) 

a,nd  so  on ;  and  the  required  sum  may  be  obtained  from  the 
known  sum  of  the  series 

[cos  ma  +  cos  m  (« -f-  )3)  +  cos  m  (a  +  2/3)  H ] 

and       I  cos  (m  —  2)  «  +  cos  [  (m  —  2)  (u  +  /8)  ] 

•  +cos[(m-2)(«  +  2y3)]  +  -..|,  etc. 

We  may  find  the  sum  of  the  scries 

cos"*  n  +  cos"*  (a  +  iS)  +  cos"*  («  +  2  /8)  +  etc. 

to  n  terms  in  a  similar  manner  by  the  aid  of  Art.  168. 


Hence 


■> 


0. 


EXAMPLES. 

1.    Sum  to  n  terms  the  series 

sin* rt  4-  sin=»  («  +  /8)  +  sin*  (a  +  2  /3)  +  ... 

We  have 

2sin2a  =  -(cos2rt-l)by  (1), 

2  8in2(«  +  i3)  =  -  [cos  2  («  +  ^)  -  1]  by  (2), 

2  sin*  (a  +  2  ^)  =  -  [cos  2  («  +  2  y8)  -  1],  and  so  on. 

Hence 

2S„=n— [co8  2rt+co8  2(rt+/8)+cos2(«+2/3)  +  ...] 

^^^cos[2«  +  (n-l)^]sinrt^  (Art.  175) 

sin  /3  ^  '^ 

•    S  —^     cos[2ot +  (n  — l)j8]sinn)3 
"     "     2  2sin)3 


254 


PLANE  TRIGONOMETRY. 


2.   Sum  to  n  terms  the  series 

cos' u  4-  cos"* 2u  +  cos' 3 «  + 


.       2  cos  ['A  /<  -f  I  (>t  —  1)  3  rt"|  sin  f  nu 
8  sin  J  « 


(i  008 


rt  H —  « 


•> 


sm 


nit 


8  sin  ^  a 
177.   Sum  the  Series 

sinu  —  sin  («;  -f  /3)4-sin  («  +  2/3) to  7i  terms 

Change  ft  into  P  +  n,  and  (1)  becomes 

8in«  +  8in(«  +  tt  -f  ;3)  +  sin(rt  +  'Jir  +  2/3)  +  ...". 

Therefore  we  liave  from  Art.  174 

+  P) 


sin 


S.= 


, -^ J - 


sin 


Z+1 


2 


raB 


•  •         • 


Similarly, 

cos«  — co8(«4-/3)  +  co8(«-f2/3) to  n 

cosr«  +  (!LzlIK5L±^"]  sin  !iOL±^ 

"  8iu2L±i:" 

2 

17a  Sum  the  Series 

cosec  6  4-  cosec  2  6^  -f-  cosec  46  -\-  •"  to  n  terms. 

We  have        cosec  tf  =  cot  -  —  cot  $, 

40 

coser  "Jtf  =  cot  tf  —  cot  2$, 
etc.  =  etc. 

cosac  2"  -'d  =  cot  2*-''$  -  cot  2="-'^. 
Therefore,  by  addition,  as  in  Art.  174, 

S^  =  cot|d-cot2'^-"tf. 


(1) 
(2) 


•         •         • 


(3) 


(4) 


SUMMATION  OF  SERIES. 


255 


Note.  —  Tho  nrtlRce  employed  in  this  Art.,  of  reBolvInK  each  term  Into  the  dif- 
ference of  tM'o  other*,  in  exlenaivcly  iiMeii  In  the  niimninlion  of  Kerie*. 

Practice  alone  will  give  the  student  readineitii  in  eHeotinK  nuvh  traniiformationa. 
If  be  cannot  diicover  the  mode  of  redoltitlon  In  any  example,  he  will  often  easily 
recognize  It  when  ho  aeea  the  result  of  summation . 

The  Htudent,  however,  la  advised  to  resort  to  this  method  of  solution  only  as  a  last 
resoarce. 


(•^) 


.     (4) 


179.  Sum  the  Series 

6  9 

tan  6+  ^  tan  -  +  \  tan    -f  ...  to  n  terms. 
J  4 

We  have  tan  tf  =  i;ot  tf  —  2  cot  2  0, 

$  6 

^ tan -=  \ cot^ —  cot $,  . 

Id 

itan-=:|cot--^cot-, 


etc  =s  etc. 

\   .       B         1       .    ^ 

- — ;  tan  — ■  =  — -  cot  — - 
2«-i       2»->     2"~'       2""' 


1       ,    ^ 

cot 


on-a 


•>»-a 


•••  S.  =  2^,cot^i-,-2cot2#. 

180.  Sum  the  Series 

8in«-f  a;8in(«4-/3)-f  jc'8in(«-|-2/8)H — af"'sin[«-}-(n— 1)^]. 

Denote  the  sum  by  S„,  and  substitute  for  the  sines  their 
exponential  values  (Art.  161).     Thus, 

2  tS,.  =  (e<«  -  -?-<»)  -f  3;(e<(»+^)  _  e-*(»+P)) 


g<«  _  a;"e<(«  +  np)      er  <«  —  aj"c-  *(" + »'') 


1-xcO 


l-(Be-<^ 


[Alg.  (3)  Art.  163] 


"^  1  -  a?(  e<^  +  e-'*P)  +  .r" 


J^ 


256 


PLANE  TRIGONOMETRY. 


.'.  S.= 


U 


sin«— X3iu(rt— ^)--j;'*8in(«-f-n/3)-i-af *^'sin[«+(n— 1)/3]    .^>. 


1  —  2xco8/8  +  «* 
Cor.   If  aj  <  1,  and  n  be  indefinitely  increased, 


S    _  si"  a  —  X  sin  («  —  ^) 
l  —  2x  cos  /3  +  «* 


(2; 


5cA.   Similarly, 

cos«  -|-  a;  cos  («  +  )3)  +  a''cos  («  +  2)3)4-  •••  to  n  terms  = 

cosft— a;co8(«— /3)— a;"co8(«-f  n/8)-f  a^+'co8[rt+(^7i  — 1)/8] 

1  —  2a;  cos /3  -j-ar* 

We  may  obtain  (3)  from  (1)  by  changing  «  to  « -f-^. 


{^) 


Also  g^^co8«-a;cj)sl«-|)    ....     (4) 

l-2«cos/8  4- X*  ^ 


181.  Sum  the  Infinite  Series 


(Bsin  («  +  ^)  +  i^sin  («  +  2y3)  +  ^sin  («  +  3)8)  -f-  •••, 
If  l^ 

and  a;co8(«  +  )3)  +  ^co8(«  +  2/3)  +  ^cos(«  +  3)3)+  •••. 

Let  S  denote  the  former  series,  and  C  the  latter. 

Then  C  +  tS  =  a;e<(<'  +  '')  +  -c*(»  +  2^)  .  ^g^Ca+sp)^.  ... 

[2  [3 

\   ^\2    ^|3    ^  ; 

=  e*«(e*«*^-l)      .     .     [by  (3)  of  Art.  129] 
=  e»«(gx(co.^  +  i«inp)_i)     .    .     .     (Art  161) 

— .  gc  COS  Pgi(a.  +  X  sin  fi) gta 

_  gx coBflj-cos (rt 4-a;  sin  )3) -f  i  sin (« +a5sin )3) ] 
—  (cos  «  + 1  sin  «)    ....     (Art.  161) 


m 


.     (4) 


EXAMPLES. 

Equating  real  and  imaginary  parts,  we  have 
C  =  e-^co*^  cos  (rt  4-  X  sin  P)  —  cos  «, 
S  =  e*<=°"^  sin  («  +  x  sin  ft)  —  sin  a. 

EXAMPLES. 

Prove  the  following  statements  : 

1.   The  two  values  of  (cos  4  ^  +  V—  1  sin  4  tf) -  are 


257 


±(eos2^  +  VI^sin2d)      .    .     .     (Art.  154) 
2.    The  three  values  of  (cos  ^  -f  V—  1  sin  d) i  are 
cos^+V^^sin^,   cos^^-HV-isin?^^^, 


a 


3.    The  three  values  of  (—1)^  are 
1_|_V~3         .     1-V^^ 


3 


(Art.  154) 


2        '         '  2 

4.  The  six  values  of  (—1)*  are  contained  in  v 

cos  i^''  +  ^)''  ±  y/-iri  sin ^^^-±-^- ,  where  r  =  0,  1,  or  2. 

5.  The  three  values  of  (1  H-V—  1)»  are  contained  in 
2* ["cos  I  +  V^^  sin  n  where  0  =  -,  f  rr,  or  V  «•■ 

B.   The  three  values  of  (3  +  4 V—  1) '  are  contained  in 
cos —J 1-  V  —  1  sin  — — -! —  ,  where  r  =  0,  1,  or  2. 

7.  cos  6  ^  =  cos"  ^  —  15  cos*  6  sin* ^4-15  cos' 0  sin*  ^-sin«  d. 

8.  sin  9$-d  COS*  tf  sin  ^ - 84  cos"  $  si n""  ^ -f  1 2()  cos*  ^  si n»  tf 

-36co8«^8iu^d-t-iiiu"^, 


V.) 


258 


PLANE  TRIGONOMETRY. 


9.    tan  n6 


II 


I 


n  tan  ^  -  n  (ti_--l )  ( u  -  2)       , 


10.   Given =  -— ^:  show  that  6  is  nearly  the  circular 

e       2166  -^ 

measure  of  3°. 


Prove  the  following : 

11.  -  64  sin^  6  =■•  sin  7  6-7  sin  5  6  -f-  21  sin  3  0  -  35  sin  0. 

12.  -.2»sin'"^=cos  10 ^Si-lOcos 8 O+in cos 6^-120 cmAe 

+  210  cos  2  (9 -126. 

13.  2«  (cos«  d  +  sin"  0)  =  cos  8  ^  +  28  cos  -l  d  +  35. 

14.  cos«0-f  siii''e=  J(54-3co84tf). 

16.   Expand  (sin  0)*"+"  in  terms  of  cosines  of  multiples 

ote. 

16.  Expand  (sin  9)*"-*-^  in  terms  of  sines  of  multiples  of  $, 

17.  Expand  (cos  ^)^"  in  terms  of  cosines  of  multiples  of  0. 

Use  the  exponential  values  of  the  sine  and  cosine  to 
prove  the  following : 

iQ        sin  ^  .$ 

18. -  =  cot-. 

1  -  cos  $  2 


19.   If  log(.T4-y  V— i)  =  rt-|-/3V-l,  prove  that 
ar*  +  y-  =  e'-",   and   y  =  x  tan  p. 


20.  If  sin (rt  + /9V— 1)=  a; 4- yV— 1,  prove  that 

a;*  co8e(;'''  u  —  if  sec* «  =  1. 

21.  2  co8(h  cos-'j;)  =  (.u  +  \A^  Vl  —  ^Y 


EXAMPLES. 


269 


-X-*)*. 


22.    (V^iy-'  =  e'". 


_  pV2(  C 


23,  e''(cos  0  +  V-  1  sin  ^)  =  eVal  cos  ^  4.  V-  1  sin  " 

24.  The  coefficients  of  x"  in  the  expansion,  (1)  of  e"  cos  bx, 
and  (2)  of  e"'  sin  bx,  in  powers  of  x,  are 

l^M:J!!)?eosn^and(^±^siun^. 

[n  l« 

25,  The  coefficient  of  x"  in  the  expansion  of  e"  cos  x  in 

.      .     2^         ?l7r 
powers  of  a;  is  —  cos 

|«  4 

26,  If  the  sides  of  a  right  triangle  are  49  and  51,  then 
the  angles  opposite  them  are  43°  51'  15"  and  46°  8'  45" 
nearly, 

27.  If  «  and  b  be  the  sides  of  a  triangle,  A  and  B  the 
opposite  angles,  then  will  log  b  —  log  a 

=cos  2  A  —  cos  2  B  +  ^  (cos  4  A  —  cos  4  B) 

+  ^ (cos  6  A  —  cos  6  B)  +  . .'.! 

28.  If  A  4-  tB  =  log(m  -f-  in),  then 

tan  B  =  --,  and  2  A  =  log  (n^  +  m^) . 


m 


29.   cos(e-f  j'<^)  =  cos^| 


''e-*  -f  e*\ 


4-  i  sin  <9 


'^e-'p  — 


30,   s'm{e-\-ii>)=sine(^-^^^^]-ico8e 


31.  2  cos(«  -f-  ift)  =  cos «(e''  +  e-'^)  -  i  sin  rt(e^  -  e-^). 

32.  (a4-i6)(»+'^> 

=  r"e -^»*[cos(j3  log r  4- rtr)  4- i sin (/?  log  r  4- «r) ], 

where   a+  ib=  r(cos  r  4  i  sin  r). 

33.  log(a  +  t6)  =  ilog(a='4^')4-t"tan->-. 


■-■'   ! 


ft. 


260 


PLANE  TRIGONOMETRY. 


■ffljlBk' 


34.  [8in(«-d)4-ct'«8intf]'' 

=:sin"~'«[sin(«  —  H^)  +  ei*«sin  n$]. 

35.  ^  =  JL_  + J_4-_L_4..... 
8     1.3     5-7     9.11 

36.  Write  down  the  quadratic  factors  of  ic"  —  1. 

Ans.    {x  —  l)[x'  —  2xcos^(2rir)'\-l],  six  factors, 
putting  r  =  1,  2,  3,  4,  6,  6. 

37.  Solve  the  equation  aJ*  —  1  =  0. 

Ans.  {x'-l){x'-x  +  l)(a^-^x-\-l)  =  0. 

38.  Give  the  general  quadratic  factor  of  ic*  —  a*. 

Ans.  mP  —  2  ax  cos  i^^^  (rTr)  -f  a^ 

39.  Find  all  the  values  of  ^/l. 

Ans.  cos  j^(r7r)4- tsin  ^(7-7r),   r  having   each  integral 
value  from  0  to  11. 

40.  Write  down  the  quadratic  factors  of  afi  +  1. 

Ans.   (x'  -  V3  ;c  4- 1)  («*  +  l)(a^  4-  V3a;  +  1). 

41.  Write  down  the  general  quadratic  factor  of  a^  +  1. 

Ans.  a^  -  2  a;  cos  (1  4-  2  r)  9"  +  1. 

42.  Find  the  factors  of  a;*^  4. 1  =  0. 

Ans.  (x-f  1)  [ar*  — 2  a;  cos  3Jjj(7r  + 2  rTr)  4- 1],  seven  fac- 
tors in  all. 


43;    Find  a  general  expression  for  all  the  values  of  >/—  1. 

Ans.  co8^^-^t — -  4-  It  sin  — "*"  ""  ^^,   where  r  may  have 
n  n 

any  integral  value. 

44.    Solve  a;"  _  2  .V*  cos  ^|  TT  -I- 1  =  0. 

Ans.  a*  —  2  a?  cos  ^  (3  >*«■  4-  tt)  4- 1  Hi  0,  six  quadratics. 


i^];.' 


EXAMPLES. 


261 


45.  Solve  »•»  4-  V3  ar*  +  1  =  0. 

Ana.  x^  +  2x  cos  {r  x  72°  +  0°)  +  1  =  0,  five  quadratics. 

46.  Write  down  the  quadratic  factors  of 

cc*"  —  2  x^y"  cos  «  -I-  2/*". 

Ans.  a^-2xycos  'L±^^J[  +  y*,  n  factors. 


n 


Prove  the  following : 

47.  tan <A  tan/'<^  +  ^Van/'*^  +  ^^Y.-tanCc^  +  ^"^^^^ 

(-  1)*,  where  n  is  even.     [Use  (2)  of  Art.  172.] 

48.  sin  5  ^  -  cos  5  ^  =  16  cos  (^  —  27°)  cos  {$  +  9°)  x 

X  sin  {0  -\-  27°)  sin  {$  -  9°)  (cos  0  -  sin  9). 

50.  ..-.-.=2«(i+5)(i+^...     ■  :>  :, 

61.   1  +  1+14.1  +  ...=='^.  .. 

(i:9l|,l_i_l|li  t' 

iii  +  3i  +  5!i  +  72+-  =  8-     -  -  «        «     '.i^ 

;.o  o   36    144   324   676  . 

35    143   323   676  ;        V 

64    !!:_,2.2.4.4.6.6.8.8...  "  '         '^ 


2     1.3.3.5.5.7-7.9... 
4.36.100.196.324... 


55.    V2  = 
66.   ^V3  = 


3.35.    99.195.323... 

8^80.224.440... 
9.  81.  225. 441  ..^ 


tV-   »' 


II) 


262 


PLANE  TRIGONOMETRY. 


67.  cos  X  -f  tan  ^  sin  a;  = 


68.  cos  X  —  cot  ^  sin  x  = 
2 


V       yA      2rr-;yA      ^Tr+yA      ^Tr-yA      4^+; 


'  +yj*" 


sin  2^ 
sin  6) 
deduce  the  value  for  cos  d  obtained  in  Art.  173 


69.   By  aid  of  the  formula  cos  $  =  ^*" "' "  and  Art.  172, 


GO.   By  expanding  both  sides  of  Ex.  57  in  powers  of  x 
and  equating  the  coefficients  of  x,  prove  that 


tan  f  = 


.V__2 2_ 


+  o 


2    ^      2 


2     v—y     ir+y     Sir—y     37r+y     Sw— y     57r+y 
61.   Prove  in  like  manner  from  Ex.  68  that 


+ 


cotV  =  2 2_  +  _J_ 2_^_A__ 

2     y     27r-y     27r  +  y     47r-y     4n--fy 

63,   Prove  ^-^^=l-»4-l-l+f3-j\  +  A 


64.  Prove  that 


»+  * 


siny 
1 


^  +.  1 


y    TT— y    27r— y     7r+y    27r+y    Srr—y    Av-y    Sn+y 
Sum  the  following  series  to  n  terras : 


+ 


*^6.   sina+sin2«4-sin3aH |-sin7ia  = 


.    »» 4- 1      .    nrt 

8in     ^    a  sin 

2  2 


sin 


« 


EXAMPLES. 


263 


66.   cos  «  4-  cos  2  rt + cos  3  «  H 1-  cos  nu  = 


•    u 
sm- 

o 

4t 


67.   8iuu  +  sino«-fsiu5«-| —  = 


sin  a 


nn 


68.   co8«4-cos3rt  +  cos5rtH —  = 


sin2na 
2  sin  a 


nn      •  2    .    •    'o     .    •  !!o     .          wsinrt— 8inn«C08(M4-t)« 
69.   smV-+sin-2«-f siir3«H —  = —-. s.:^   '   . 


2  8in« 


TA    «^„«    I  ^  .,»o     I  «^„2'j     I  n8in«-l-cos(n-f-l)«8iuw« 

70.  co8*rt4-co8-2«-f-co8't>«4-"*=  - — ^    J  — / _ . 

2aiu/« 

71.  sin'rt  +  8iu='(rt  +  /3)  +  sin3(«  -{.'2p)-\--" 

^  .4—^^Jsm^     ^s.n(^.^.4-  -^3^J8in-^/ 

~4  sin^/3  4  siii  | /i 


72.   sm''rt4-8in=»2«  +  siii^']M  + 


sill       8111  (  — '      ]u     sin         sin  ~  v  - ' — '— 
3        2        V     2     y  2  2 


sm 


2 


.    •    3  u 
4  8m  ^. 


73.  8ina8m2«  +  sin2rt8iu3«  +  sin3«  8in4«  +  •••        • 

—  MsinKCosK  —  sin  n«  cos  (n  +  2>« 
2sin« 

74.  tan  «  +  2  tan  2  «  +  2«  tan  2"  «  +  •••  =  cot «  -  2"  cot  2"  u. 

75.  (tan  «  4-  cot  a)  +  (tan  2  a  +  cot  2  a)  +  (tan  2*«  -f-  cot  2^ a) 

H =2  cot «  —  2  cot  2" «. 

76.  secasec  2rt  4- sec  2«sec3rt  4- ••• 

=  coseu  u  [tail  (71  -f  1)  «  —  tan  wj . 


264 


PLANE  TRIOONOMETRT. 


:*' 


I 


77.  cosec  a  cosec  2  a  +  coseo  U  a  cosec  3  « -|-  ■ . . 

=  coseo  a  [cot «  —  cot  (n  +  1 )  «]. 

78    8in2tf   ,    sin  4  $        ....__  8ec(2n  +  l)^-8ec  ^ 
cos  $  cos  3  0  cos  3  6  cos  5  ^  2  sin  0 

79.  cos* a  +  008*  (rt  +  j3)  +  cos*  («  +  2 /?)  -f  ...  =  f  7i  + 

C08[2ft+(n-l)/3]8in7i^     cos  [4«  +  (»-l)2^]sin  2n/8 


2sin/3 


+ 


8  8in2y3 


80.   tanwtf  = 


sin  ^  4-  sin  3  ^  +  sin  5  tf  +  •  •  •  to  n  terms 


coH 6  +  COS 3 ^ -f  008 CtO  -\-  •'•to  n  terms 

81.   COS  $  cos  {6  +  a)  +  cos  (d  +  «)  cos  (d  +  2  «) 

+  cos(tf  H-  2  rt)cos(e  -I-  3  «)  H 

—  !^ cos  «  4-  P08(2^  +  ^«)^"'^« 
2  2  sin  n 

on    sin  0— sin  2d4-8in  3tf to  n  terms     .      n  +  1.     ,  a\ 

82. — =  tan       —  (Tr-ft'). 

cos  tf — cos  2  tf -j-oos  3  $~ ...  to  n  terms  2 

83.  sin(p  +  l)tfcosd  +  sin(/)  +  2)dcos2e4---- 

__  n  sinjj^  ,  sin  (jj  4- 1  +  n)6  sin  n$^ 
"2  2sind 

84.  sin  3  6  sin  d  +  sin  G  tf  sin  2  d  +  sin  12  ^  sin  4  ^  +  •  •  • 

=  ^(oos2^-eos2"+>d). 


8*  sintf 


fs'm  ^Y+  2  sin  -("sin  -Y+  4  sin 


e 


=  2"--  sin  --"^  -  V  sin  2  ^. 


0  BOBS  9 

86.  tan    sec^  + tan-sec- 4- tan-sec   -f...=tand— tan— . 

2  4       2  8       4  2" 

87.  cot  6 cosec  ^+2  cot  2  tf  cosec  2  ^+2"  cot  2^6  cosec  2''^+ ... 

1 


On-l 


2  sin 


2^     sin=*2"-^d 
2 


EX  A  MPLES. 


265 


tan 


I 

2"' 


88. 


sin 


_1 +  ____! .  1  . 

e  sin  2  $     sill  2  6  sin  ;J  $     sin  3  ^  sin  4  ^ 


--v,(cot3tf-cot4^). 
sm  ^ 


89. 


TTTTi-^ZT. 


sin  tf  cos  2  0     cos  2  tf  sin  3  tf     sin  3  ^  cos  4  6 
=  cosec(d  +  ^ 


)  tan(n4-l)(^  +  ^Vtai/^  +  ^y 

90.   tau-i  ^ 1— --  +  tan-' 1 +  tan-» ^ 

1  +  1  +  1'  1  +  2  +  2"^  l+3  +  3« 


+  ...  =  ^--tan-'-i~. 
4  w  +  1 


91.   tan-'  X  +  tan"' 


X 


l  +  1.2-a:« 


+  tan"' 


X 


;  + 


1  +  2  . 3  .  a?« 

=  tan-'  nx. 


92.    sin  «  sin  3 «  +  sin" sin '"^^  +  sin  "  sin—  +  ... 

2        2  2*       2* 


=    (  cos    COS  4 

2 1       2"  =» 


"> 


93. 


;  + 


+ 


+ 


cos  ^  +  COS  3  ^     cos  ^  +  cos  5  0     cos  ^  +  cos  7  tf 

=  ^  cosec  ^[tan(n  +  1)0  —  ta)i  6]. 

94.  -sec^  +  -sec^sec2^  +  ,^-8ec^sec2esec22^+  ... 

=  sin^(cot^-cot2"^). 

95.  ^log  tan  2  tf  +  |^log  tan  2^^  + 1  log  tan  2«^  +  ... 

=  log 2  sin  2$--  log 2 sin  2"^ '  tf. 
Sum  the  following  series  to  infinity : 

96.  cos^+«"^-^cos2^+5^^cos3^+^«fi^cos4tf+... 

1  [L  [3 

—  eco.»«  cos(^  +  sin  $  cos  $). 


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I 


I 


IP' 


266 


PLANE  TRIGONOMETRY. 


9/.  sin^ — 1 — — =  e-cos*  sin(sin^). 

98.  1  _  ^J^^  +  22ii^ =  icos(cos 6)  (e«i"«  +  e-«'°»). 

99.  2cos^  +  fcos2^-hf  cos3e  +  fcos^^  +  ...  ;V-,  ' 

cos^ 


1  —  COS  0 


—  log(l  —  cos^). 


mn     r,-    a        a  ,  sin2^cos^^  ,  sinS^cos^^  , 
100.    sin  d  COS  Q  H ■ 1 , 1- 


I? 


[3 

_  eco8«0sin(sin^cos^). 


101.  cos^  +  ?H^cos26  +  ^^cos3^4--" 

.        ,1.  [2 

=-e8iQflco8  0  cos(^  +  sin^e). 

102.  sin^+-— sin2^  +  ^-^'^-sin3^4- 

<    \  =  e«'n^  <=»''«  sin  (^  +  sin2^). 

103.  cos^-|cos2^4-^cos3^ =log/'2cos^Y      . 

104.  cos  2^+^  cos  65  + I  cos  10^4-  •••  =  |log  cot-. 

M\r         -a    a^sin2^  ,  ar'sin3^  ..i/cosec^  ,       .A 

105.  ifsm^ 1 =  cot  '( f-cot^l- 

2  3  \     X  ) 

T  Oj  or 

106.  a;  cos  6  —  '—  cos  '26  -\ —  cos  3$ cos  4  ^  +  •  •  • 

2  3  4 

.       f  =  log(l  + 2a;  cos  ^  +  ar'). 

107.  sin^      sin2^ hsin3^ ••• 

1  2  3 

=  cot-' (1  +  cot^  ^  4- cot  ^) . 

1  1  1  1  -4 


1*      2*     3^     44 


90 


14-^-34-^54-^74-1-  9g 


,-Blnfl> 


-COS^). 


/  ■  / 


OcosB). 


l-siii^e). 


f  siii^^). 


+cotd 


)S^4-arO- 


?  +  cotd). 


PART  IT. 


SPHERICAL    TRIGONOMETRY. 


CHAPTER   X. 

rORMTJLU  KELATIVE  TO  SPHERICAL  TRIANGLES. 

■  :  182.  Spherical  Trigonometry  has  for  its  object  the  sohi- 
tion  of  spherical  triaiigles.  .     '     ' 

A  spherical  triangle  is  the  figure  formed  by  joining  any 
three  points  on  the  surface  of  a  sphere  by  arcs  of  great 
circles.  The  three  points  are  called  the  vertices  of  the 
triangle ;  the  three  arcs  are  called  the  sides  of  the  triangle. 

Any  two  points  on  the  surface  of  a  sphere  can  be  joined 
by  two  distinct  arcs,  which  together  make  up  a  great 
circle  passing  through  the  points.  Hence,  when  the  points 
are  not  diametrically  opposite,  these  arcs  are  unequal,  one 
of  them  being  less,  the  other  greater,  than  180°.  It  is  not 
necessary  to  consider  triangles  in  which  a  side  is  greater 
than  180°,  since  we  may  always  replace  such  a  side  by  the 
remaining  arc  of  the  great  circle  to  which  it  belongs. 

183.  Geometric  Principles.  —  It  is  shown  in  geometry 
(Art.  702),  that  if  the  vertex  of  a  triedral  angle  is  made 
the  centre  of  a  sphere,  then  the  planes  which  form  the 
triedral  angle  will  cut  the  surface  of  the  sphere  in  three 
arcs  of  great  circles,  forming  a  spherical  triangle. 

Thus,  let  0  be  the  vertex  of  a  triedral  angle,  and  AOB, 
BOC,  COA  its  face-angles.  We  may  construct  a  sphere 
with  its  centre  at  O,  and  with  any  radius  OA.     Let   AB, 

267 


268 


SPHERICAL    TRIGONOMETRY, 


BC,  CA  be  the  arcs  of  great  circles  in  which  the  planes  of 

the  face-angles  AOB,  BOC,  COA 

cut  the  surface  of  this  sphere; 

then  ABC  is  a  spherical  triangle, 

and  the  arcs  AB,  BC,  CA  are  its 

sides. 

Now  it  is  shown  in  geometry 
that  the  three  face-angles  AOB, 
BOC,  COA  are  measured  by  the  sides  AB,  BC,  CA,  re- 
spectively, of  the  spherical  triangle,  and  that  the  diedral 
angles  OA,  OB,  OC  are  equal  to  the  angles  A,  B,  C,  respect- 
ively, of  the  spherical  triangle  ABC,  and  also  that  a  diedral 
angle  is  measured  by  its  plane  angle. 

There  is  then  a  correspondence  between  the  triedral 
angle  0-ABC  and  the  spherical  triangle  ABC :  the  six 
parts  of  the  triedral  angle  are  represented  by  the  corre- 
sponding six  parts  of  the  spherical  triangle,  and  all  the 
relations  among  the  parts  of  the  former  are  the  same  as 
the  relations  among  the  corresponding  parts  of  the  latter. 

184.  Fundamental  Definitions  and  Properties.  —  The  fol- 
lowing definitions  and  properties  are  from  Geometry,  Book 
VIII.  : 

In  every  spherical  triangle 

Each  side  is  less  than  the  sum  of  the  other  two. 

The  sum  of  the  three  sides  lies  between  0°  and  360°. 

The  sum  of  the  three  angles  lies  between  180°  and  540°. 

Each  angle  is  greater  than  the  difference  between  180° 
and  the  sum  of  the  other  two. 

If  two  sides  are  equal,  the  angles  opposite  them  are 
equal ;  and  conversely. 

If  two  sides  are  unequal,  the  greater  side  lies  opposite 
the  greater  angle  ;  and  conversely. 

The  perpendicular  from  the  vertex  to  the  base  of  an 
isosceles  triangle  bisects  both  the  vertical  angle  and  the 
base. 


/ 


DEFINITIONS  AND  PROPERTIES. 


269 


,  CA,  re- 

e  diedral 

),  respect- 

a  diedral 

^  triedral 
:  the  six 
the  corre- 
d  all  the 
3  same  as 
3  latter. 

-The  fol- 
try,  Book 


360°. 
lud  540°. 
ween  180° 

them   are 

s  opposite 

ase  of  an 
e  and  the     ^ 


The  axis  of  a  circle  is  the  diameter  of  the  sphere  perpen- 
dicular to  the  plane  of  the  circle.  The  poles  of  a  circle  are 
the  two  points  in  which  its  axis  meets  the  surface  of  the 
sphere. 

One  spherical  triangle  is  called  the  polar  triangle  of  a 
second  spherical  triangle  when  the  sides  of  the  first  triangle 
have  their  poles  at  the  vertices  of  the  second. 

If  the  first  of  two  spherical  triangles  is  the  polar  triangle 
of  the  second,  then  the  second  is  the  polar  triangle  of  the 
first. 

Two  such  triangles  are  said  to  be  polar  with  respect  to 
each  other.     Thus: 

If  A'B'C  is  the  polar  triangle  of 
ABC,  then  ABC  is  the  polar  triangle 
of  A'B'C. 

In  two  polar  triangles,  each  angle 
of  one  is  measured  by  the  supplement 
of  the  corresponding  side  of  the  other. 

Thus : 

A  =  180°  -  a',         B  =  180°  -  6',         C  =  180°  -  c', 
a  =  180° -A',         6  =  180°-B',        c  =  180°-C'. 

This  result  is  of  great  importance ;  for  if  any  general 
equation  be  established  betAveen  the  sides  and  angles  of  a 
spherical  triangle,  it  holds  of  course  for  the  polar  triangle 
also.  Hence,  hy  means  of  the  above  formidoe  any  theorem  of 
a  spherical  triangle  may  he  at  once  transformed  into  another 
theorem  hy  suhstituting  for  each  side  and  angle  respectively 
the  supplements  of  its  opposite  angle  and  side. 

If  a  spherical  triangle  has  one  right  angle,  it  is  called  a 
right  triangle ;  if  it  has  two  right  angles,  it  is  called  a  bi- 
rectaMgular  trisingle ;  and  if  it  has  three  right  angles,*  it  is 
called  a  tri-rectangular  triangle.  If  it  has  one  side  equal  to 
a  quadrant,  it  is  called  a  quadrantal  triangle ;  and  if  it  has 
two  sides  equal  to  a  quadrant,  it  is  called  a  bi-quadrantal 
triangle.  ^  ^^  ? 


i 


Ill  ^f 


I'    "J  I 


liiil'  r 


270 


SPHERICAL    TRIGONOMETRY. 


Note.  —  It  is  shown  in  geometry  that  a  spherical  trianirle  mny,  in  general,  be 
constructed  when  any  three  of  its  six  parts  are  given  (not  excepting  the  case  in 
which  the  given  parts  are  the  three  angles).  In  spherical  trigonometry  we  investi- 
gate the  methods  by  which  the  unknown  parts  of  a  spherical  triangle  may  bo  com- 
jfutcd  from  the  above  data. 

EXAMPLES. 

1.  In  the  spherical  triangle  whose  angles  are  A,  B^  C, 

prove 

B+C-A<^ (1) 

C+A-B<7r (2) 

A  +  B-C<7r (3) 

2.  If  C  is  a  right  angle,  prove 

A  +  B<f7r  (1),  andA-?,<J  (2). 

3.  The  angles  of  a  triangle  are  A,  45°^  and  120°;  find  the 
maximum  value  of  A.  Ans.  A  <  105°. 

4.  The  angles  of  a  triangle  are  A,  30°,  and  150° ;  find  the 
maximum  value  of  A.  Ans.  A  <  60°. 

5.  The  angles  of  a  triangle  are  A,  20°,  and  110° ;  find  the 
maximum  value  of  A.  Ans.  A  <  90°. 

6.  Any  side  of  a  triangle  is  greater  than  the  difference 
between  the  other  two. 

RIGHT  SPHERICAL  TRIANGLES. 


185.  Formiilse  for  Right   Triangles 

spherical  triangle  in  which  C  is 
a  right  angle,  and  let  0  be  the 
centre  of  the  sphere ;  then  will 
OA,  OB,  OC  be  radii:  let  a,  b,  c 
denote  the  sides  of  the  triangle 
opposite  the  angles  A,  B,  C,  re- 
spectively ;  then  a,  b,  and  c  are 
the  measures  of  the  angles  BOC, 
COA,  and  AOB. 


Let    ABC    be    a 


RIGHT  SPHERICAL   TRIANGLES. 


271 


neral,  be 
e  case  in 
e  invcBll- 
f  bo  com- 


,  B,  C, 


find  the 
\.  <  105°. 

,  find  the 
A  <  60°. 

;  find  the 
A<90°. 

difference 


From  any  point  D  in  OA  draw  I)E  _L  to  OC,  and  from  E 
draw  E F  ±  to  OB,  and  join  DE.  Then  DE  is  ±  to  EE  (Geom. 
Art.  537).     Hence  (Geom.  Art.  507), 

DE  is  J_  to  OB  ;    .-.  Z  DEE  =  Z  B  .     .     (Art.  183) 


,T       OE      OE    OE     ,,    ,  .  , 

Now  -7^  =  — rr  •  -r-zr]  that  IS,  COS  c  =  COS  a  cos  0 


OD      OE    OD 

DE^DE   DE 
OD 


DE   OD 

Interchanging  a's  and  6's, 


;  that  is,  sin  b  =  sin  B  sin  c 
sin  a  =  sin  A  sin  c 


EE     EE    DE    , ,    ,  .     ,  T,  , 

—  =  —  •  — ;  that  IS,  tan  a=  cos  B  tan  c  . 

OE     DF    OE'  ' 

Interchanging  a's  and  &'s,        tan  b  =  oos  A  tan  c  . 

DE      DE    EE     .,    ,.     .      ,       .      ^    . 
= ;  that  is,  tan  b  =  tan  B  sin  a  . 

OE      EE    OE'  ' 

Interchanging  a's  and  &'s,        tan  a  =  tan  A  sin  b  . 
Multiply  (6)  and  (7)  together,  and  we  get 


tan  A  tan  B  = 


cos  a  cos  b      cos  c 
•.  cos  c  =  cot  A  cot  B 


=  -i-,hy(l) 


Multiply  crosswise  (3)  and  (4),  and  we  get 

sin  a  cos  B  tan  c  =  tan  a  sin  A  sin  c. 

-r,      sin  A  cose        .     .         j,  i,     /i\ 
.-.  cosB  = =sinAcoso,  by  (1)    . 

cos  a 
Interchanging  a's  and  6's, 

cos  A  =  sin  B  cos  a  .     .     .     . 


(1) 

(2) 
(3) 
(4) 

(5) 
(6) 


(8) 


(9) 


(10) 


Sch.  By  these  ten  formulae,  every  case  of  right  triangles 
can  be  solved ;  for  every  one  of  these  ten  formulee  is  a  dis- 
tinct combination,  involving  three  out  of  the  five  quantities, 
a,  b,  c.  A,  B,  and  there  can  be  but  ten  combinations  in  all. 
Hence,  any  two  of  the  five  quantities  being  given  and  a 
third  required,  that  third  quantity  may  be  determined  by 
some  one  of  the  above  ten  formulaj. 


272 


SPHERICAL   TRIGONOMETRY. 


186.  Napier's  Eules.  — The  ten  preceding  formulae,  which 
may  be  found  difficult  to  remember,  have  been  included 
under  two  simple  rules,  called  after  their  inventor,  Napier^ s 
Rules  of  the  Circular  Parts. 

Let  ABC  be  a  right  spherical  triangle.  Omit  the  right 
angle  C.     Then  the  two  sides  a 

and  •  &,   which  include  the  right  *yv 

angle,   the    complement    of    thq    • ,    .^      ^  ■/     \ 

hypotenuse   c^    and   the    comply '''*^^  <    Q.^y^  V 

ments  of  the  oblique  angles  A  jy^  \ 

and  B,  are  called  the  circular  parts      '  *=^^^C!x 'C 

of  the  triangle.     Thus,  there  are 

Jive  circular  parts,  arranged  in  the  figure  in  the  following 

order :  a,  b,  co.  A,  co.  c,  co.  B. 

Any  one  of  these  five  parts  may  be  selected  and  called 
the  middle  part;  then  the  two  parts  next  to  it  are  called 
adjacent  parts,  and  the  remaining  two  parts  are  called  opjw- 
site  parts.  Thus,  if  co.  A  is  selected  as  the  middle  part, 
then  h  and  co.  c  are  the  adjacent  parts,  and  a  and  co.  B  are 
the  opposite  parts. 

Then  Napier's  Eules  are: 

(1)  Tlie  sine  of  the  middle  part  equals  the  product  of  the 
tangents  of  the  adjacent  parts. 

(2)  The  sine  of  the  middle  part  equals  the  product  of  the 
cosines  of  the  opposite  parts. 

Note  1.  —  It  vlll  assist  the  student  In  remembering  these  rules  to  notice  the 
occurrence  of  the  vov^'el  i  in  sine  and  middle,  of  the  vowel  a  in  tangent  and  adjacent, 
and  of  the  vowel  o  in  cosine  and  opposite. 

Napier's  Rules*  may  be  made  evident  by  taking  in  detail  eacn  of  the  five  parts  ae 
middle  part,  and  comparing  the  equations  thus  found  with  the  formulsa  of  Art.  186. 

Thus,  let  CO.  c  be  the  middle  part.    The  rules  give 

Bin(co.  c)  =  tan(co.  A)  tan(co.  B);  .•.  cos  c  =cot  A  cat  B (8)  • 

Bin  (CO.  c)=  cos  «  cos  i;  .*.  cose  =  cos  a  cos  6 (1)- 

co.  B  the  middle  part. 

Bin(co.  B)=  tan  a  tan  (CO.  c) ;  .'.  cos  B  ==  tan  a  cot  c (4)d 

8in(co.B)  =  coB  b  cos(co.  A) ;  .•.  cos  B  =  cos  6  sin  A (9)" 

*  While  Bome  find  these  ruleB  to  be  useful  aids  to  the  memory,  others  qaeation 
tbeir  utility. 


THE  SPECIES  OF  THE  PARTS. 


278 


,  which 
\clude(l 
Napier'' s 


jllowing 

d  called 
re  called 
led  oppo- 
dle  part, 
co.B  are 


ict  of  the 
ict  of  the 


to  notice  the 
iiid  adjacent, 

five  parts  ae 
of  Art.  18&. 


(8)- 
(D- 

(4)- 
(9)' 


,heri  quoition 


a  the  middle  part. 

■in  a  =  tari  6  tan  (CO.  B);  .■.  Bin  a  =tan  b  cot  B .    (0) 

sin  a  =  coB(co.  A)co8(co.  c),     .-.  sin  a  ^sin  A  ain  c (3)' 

6  the  middle  part. 

Bin  6  =  tan  a  tan(co.  A) ;  .-.  sin  2)  =  tan  a  cot  A (7) 

Bin  6  =  co8(co.  c)cob(co.  B):      .■.  Bin  ft  =Bin  c  sin  B (2) 

CO.  A  the  middle  part. 

flin(co.  A)=  tan  6  tan(co.  c);     .•.  cob  A=  tan  6  cot  c (5)- 

Bin(co.  A)=coB  a  coB(co.  B) ;    .•.  cob  A=co8  n  Bin  B (W)^ 

Note  2.  —  In  applying  these  rules  it  is  not  necessary  to  use  the  notation  co.  c, 
CO.  A,  CO.  B,  since  we  may  write  at  once  cob  c  for  sin  (co.  c),  etc. 

187.  The  Species  of  the  Parts.  —  If  two  parts  of  a  spheri- 
cal triangle  are  either  both  less  than  90°  or  both  greater  than 
90°,  they  are  said  to  be  of  the  same  species.  But  if  one  part 
is  less  than  90°  and  the  other  part  is  greater  than  90°,  they 
are  of  different  species. 

In  order  to  determine  whether  the  required  parts  are  less 
or  greater  than  90°,  it  will  be  necessary  carefully  to  observe 
their  algebraic  signs.  If  the  required  part  is  determined 
by  means  of  its  cosine,  tangent,  or  cotangent,  the  alge- 
braic sign  of  the  result  will  show  whether  it  is  less  or 
greater  than  90°.  But  when  a  required  part  is  found  In 
terms  of  its  sine,  it  will  be  ambiguous,  since  the  sines  aivB 
positive  in  both  the  first  and  second  quaxirants.  This 
ambiguity,  however,  may  generally  be  removed  by  either 
of  the  following  principles : 

(1)  In  a  right  spherical  triangle,  either  of  the  sides  con- 
taining the  right  triangle  is  of  the  same  species  as  the  opposite 
angle. 

(2)  The  three  sides  of  a  right  spherical  triangle  (omitting 
bi-rectangular  or  tri-rectangular  triangles)  are  either  all 
acute,  or  else  one  is  acute  and  the  other  two  obtuse. 

The  first  follows  from  the  equation 

cos  A  =  cos  a  sin  B, 


\ 


hA^'wrn 


!*     1.     I? 


'i      i 


274 


SPHERICAL   TRIGONOMETRY. 


in  which,  since  sinB  is  always  positive  (B<  180°),  cos  A 
and  cos  a  must  have  the  same  sign ;  i.e.,  A.  and  a  must  be 
either  both  <  or  both  >  90°. 

The  second  follows  from  the  equation 

cos  c  =  cos  a  cos  b. 

188.  Ambiguous  Solution.  —  When  the  given  parts  of  a 
right  triangle  are  a  side  and  its  opposite  angle,  the  triangle 
cannot  be  determined. 

For  two  right  spherical  triangles  ABC,  X.'BC,  right 
angled  at  C,  may  always  be 
found,  having  the  angles  A 
and  A'  equal,  and  BC,  the 
side  opposite  these  angles, 
the  same  in  both  triangles, 
but  the  remaining  sides,  AB,  AO,  and  the  remaining  angle 
ABC  of  the  one  triangle  are  the  supplements  of  the  re- 
maining sides  A'B,  A'C,  and  the  remaining  angle  A'BC  of 
the  other  triangle.  It  is  therefore  ambiguous  whether 
ABC  or  A'BC  be  the  triangle  required. 

This  ambiguity  will  aldO  be  found  to  exist,  if  it  be 
attempted  to  determine  the  triangle  by  the  equation 

sin  6  =  tan  a  cot  A, 

since  it  cannot  be  determined  from  this  equation  whether 
the  side  AC  is  to  be  taken  or  its  supplement  A'C. 

189.  Quadrantal  Triangles.  —  The  polar  triangle  of  a 
right  triangle  has  one  side  a  quadrant,  and  is  therefore 
a  quadrantal  triangle  (Art.  184).  The  formukc  for  quad- 
rantal  triangles  may  be  obtained  by  applying  the  ten 
formulae  of  Art.  185  to  the  polar  triangle.  They  are  as 
follows,  c  being  the  quadrantal  side : 

cos  C  =  —  cos  A  cos  B (1) 

sin  B  =  sin  6  sin  C (2) 


(1) 

(2) 


EXAMPLES.  275 

sin  A  =  sin  a  sin  C /3\ 

cos  6  =  — tan  A  cote (4\ 

cos  a  =  —  tan  B  cot  C .     .     .    ,  /5\ 

sin  A  =  tan  B  cot  6 /g\ 

sin  B  =  tan  A  cot  a /j\ 

cosC  =  — cotacot& /g\ 

cos  b  =  cos  B  sin  a  ......     .          ....  (9) 

cos  a  =  cos  A  sin  6 /jq\ 

EXAMPLES. 

In  the  right  triangle  ABC  in  which  the  angle  C  is  the 
right  angle,  prove  the  following  relations  : 

1.  sin" a  +  sin^  b  —  sin-  c  =  sin^a  sin^b. 

2.  cos^A  sin^c  =  sin^c  —  sin^a. 

3.  sin^ A  cos^c  =  sin^  A  —  sin^a. 

4.  sin^ A  cos^d  sin^c  =  sin^c  —  sin^ft. 

5.  2cosc  =  cos(a  +  &)  +  cos(a  — 6). 

6.  tanj(c  +  a)tai4(c-a)=tan2^6. 

7.  sin2^  =  sin2^cos«^  +  cos2-sin''^. 

2  2        2  2        2 

8.  sin(c-6)  =  tan='Asin(c-H6). 

9.  If  &  =  c  =  ^,  prove  cos  a  =  cos  A. 

10.  If  a  =  6  =  c,  prove  sec  A  =  1  +  sec  a. 

11.  If  c  <  90°,  show  that  a  and  b  are  of  the  same  species. 

12.  If  c  >  90°,  a  and  6  are  of  different  species. 

13.  A  side  and  the  hypotenuse  are  of  the  same  or  oppo- 
site species,  according  as  the  included  angle  <,  or  >  -. 

2 


wr 


^E 


[■iih 


276 


SPHERICAL   TRIGONOMETRY. 


OBLIQUE  SPHERICAL  TRIANGLES. 


190.  Law  of  Sines.  —  In  any  spherical  triangle  the  sines 
of  the  sides  are  proportional  to  the  sines  of  the  opposite  angles. 

Let  ABC  be  a  spherical  triangle,  0  the  centre  of  the 
sphere ;  and  let  a,  b,  c  denote  the 
sides  of  the  triangle  opposite  the 
angles  A,  B,  C,  respectively.  Then 
a,  b,  and  c  are  the  measures  of  the 
angles  BOG,  COA,  and  AOB. 

From  any  point  D  in  OA  draw 
DG  ±  to  the  plane  BOC,  and  from 
G  draw  GE,  GF  ±  to  OB,  OC. 
Join  DE,  DF,  and  GO.  Then  DG 
is  ±  to  GE,  GF,  and  GO  (Geom.  Art.  487).  Hence,  DE  is 
±  to  OB,  and  DF  ±  to  OC  (Geom.  Art.  507). 

.-.  Z  DEG  =  Z  B,  and  Z  DFG  =  Z  C  .     .  (Art.  183) 

In  the  right  plane  triangles  DGE,  DGF,  ODE,  ODF, 
DG  =  DE  sin  B  =  OD  sin  DOE  sin  B  =  OD  sin  c  sin  B, 
DG  =  DF  sin  C  =  OD  sin  DOF  sin  C  =  OD  sin  b  sin  C. 

.'.  sin  c  sin  B  =  sin  b  sin  C ; 
or  sin  6  :  sin  c : :  sin  B  :  sin  C. 

Similarly,  it  may  be  shown  that 

sin  a :  sin  c : :  sin  A  :  sin  C. 

sin  a  _  sin  6  _  sin  c 
sin  A     sin  B      sin  C 

Note.  — The  common  value  of  theae  three  ratios  is  called  the  modulus  of  the 
spherical  triangle. 

Sch.  In  the  figure,  B,  C,  6,  c  are  each  less  than  a  right 
angle ;  but  it  will  be  found  on  examination  that  the  proof 
will  hold  when  the  figure  is  modified  to  meet  any  case 
which  can  occur.     For  example,  if  B  alone  is  greater  than 


LAW  OF  COSINES. 


277 


90°,  the  point  G  v/ill  fall  outside  of  OB  instead  of  between 
OB  and  OC.  Then  DEG  will  be  the  supplement  of  B,  and 
thus  we  shall  still  have  sin  DEG  =  sin  B. 

191.  Law  of  Cosines.  —  In  any  spherical  triangle,  the 
cosine  of  each  side  is  equal  to  the  product  of  the  cosines  of  the 
other  two  sides,  plus  the  product  of  the  sines  of  those  sides 
into  the  cosine  of  their  included  angle. 

Let  ABC  be  a  spherical  triangle,  0  the  centre  of  the 
sphere,  and  a,  h,  c  the  sides  of 
the  triangle  opposite  the  angles 
A,  B,  C,  respectively.     Then 

a  ==  Z  BOG, 
b  =  Z  COA, 
c  =  Z  AOB. 

From  any  point  D  in  OA  draw,  in  the  planes  AOB, 
AOC,  respectively,  the  lines  DE,  DF  ±  to  OA.     Then 

ZEDF  =  ZA .     (Art.  183) 

Join  EF;  then  in  the  plane  triangles  EOF,  EDF,  we 
have 


EF''=:OE'  +  OF'-2  0E.OFcosEOF  . 


r>2 


EF'  =  DE'  H-  DF'  -  2  DE  .  DF  cos  EDF 
also  in  the  right  triangles  EOD,  FOD,  we  have 


tTt]^2 


FC2 


0E'-DE'  =  0D;  and  OF  -  DF  =  OD 


(1) 
(2) 

(3) 


Subtracting   (2)   from    (1),  and   reducing  by    (3),  and 
transposing,  we  get 


or 


2  OE  .  OF  cos  EOF  =  2  OD'  +  2  DE  •  DF  cos  EDF. 

OD    OD  ,  DF    DE        ^^.j. 

.'.  cos  EOF  =  — —  •  — —  -f  — —  .  — —  cos  EDF, 

OF    OE      OF    OE  ' 

cos  a  =  cos  6  cos  c  -f  sin  &  sin  c  cos  A     (4) 


\ 


I 

ii  f  ■ 


*'S 


278 


SPHERICAL  TRIGONOMETRY. 


By  treating  the  other  edges  in  order  iii  the  same  way, 
or  by  advancing  letters  (see  Note,  Art.  96)  we  get 


cos  h  =  cos  c  cos  a  +  sin  c  sin  a  cos  B 
cos  c  =  cos  a  cos  h  -\-  sin  a  sin  6  cos  C 


(5) 

(6) 


^c^.  Formula  (4)  has  been  proved  only  for  the  case  in 
which  the  sides  h  and  c  are  less  than  quadrants;  but  it 
may  be  shown  to  be  true  when  these  sides  are  not  less  than 
quadrants,  as  follows : 

(1)  Suppose  c  is  greater  a, — <:~- ~*f.^_^ 

than  90°.     Produce  B A,  BC  B  ^^^         ^*         ~^ 
to    meet    in    B',    and    put 
AB'=c',  CB'=a'. 


>B' 


'0' 


Then,  from  the  triangle  AB'C,  we  have  by  (4) 
cos  a'  =  cos  h  cps  c'  +  sin  b  sin  c'  cos  B'AC, 
or  cos(7r— a)=:cos6cos(7r— c)  +  sin  6  sin(7r  —  c)cos(7r— A). 
.*.  cos  a  =  cos  b  cos  c  -f-  sin  6  sin  c  cos  A. 

(2)  Suppose  both  6  and  c  to 
be  greater  than  90°.    Produce     .' , 
AB,  AC  to  meet  in  A',  and  put 
A'B  =  c',  A'C  =  b'. 

Then,  from  the  triangle  A'BC,  we  have  by  (4) 
cos  a  =  cos  y  cos  c'  +  sin  V  sin  c'  cos  A' ; 


but 


6'  =  TT  —  6,  c'  =  TT  —  c,  A'  =  A. 


.*.  cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  A. 
The  triangle  AB'C  is  called  the  colunar  triangle  of  ABC. 

192.  Relation  between  a  Side  and  the  Three  Angles.  - 
In  any  spherical  triangle  ABC, 

cos  A  =  —  cos  B  cos  C  +  sin  B  sin  C  cos  a. 


-B' 


RELATION  BETWEEN  SIDE  AND  ANGLES.      279 

Let  A'B'C  be  the  polar  triangle  of  ABC,  and  denote  its 
angles  and  sides  by  A',  B',  C,  a',  b',  c' ;  then  we  have  by 
(4)  of  Art.  191 

cos  a'  =  cos  b'  cos  c'  +  sin  b'  sin  c'  cos  A' ; 
but       a'  =  TT  -  A,  6'  =  TT  -  B,  c'  =  TT  -  C,  etc.      .  (Art.  184) 
Hence,  substituting,  we  get  v 

cos  A  =  —  cos  B  cos  C  +  sin  B  sin  C  cos  a    .     .     .     .     (1) 

Similarly,  --      - 


cos  B  =  —  cos  C  cos  A  +  sin  C  sin  A  cos  b 
cos  C  =  —  cos  A  cos  B  +  sin  A  sin  B  cos  c 


(2) 


Hem.  —  This  process  is  called  "  applying  the  formula  to  the  polar  triangle."  By 
means  of  the  polar  triangle,  any  formula  of  a  spherical  triangle  may  be  immediately 
transformed  into  another,  in  which  angles  take  the  place  of  side?,  and  sides  of  angles. 

193.  To  show  that  in  a  spherical  triangle  ABC, 

cot  a  sin  6  =  cot  A  sin  C  +  cos  C  cos  b.    '  ' 

Multiply  (6)  of  Art.  191  by  cos  b,  and  substitute  the 
result  in  (4)  of  Art.  191,  and  we  get 

cos  a  =  cos  a  cos'^  b  +  sin  a  sin  6  cos  b  cos  C  +  sin  b  sin  c  cos  A. 

Transpose  cos  a  cos^  b,  and  divide  by  sin  a  sin  6 ;  thus, 

.       .    ,             ,         r\  ,  sin  c  cos  A 
cot  a  smb~  cos  b  cos  C  H 

sin  a 
=  cos  b  cos  C  +  cot  A  sin  C    .    (by  Art.  190) 

By  interchanging  the  letters,  we  obtain  five  other  formulae 
like  the  preceding  one.     The  six  formul&e  are  as  follows  : 

cot  a  sin  &  =  cot  A  sin  C  +  cos  C  cos  6      .     ...     (1) 


cot  a  sin  c  =  cot  A  sin  B  +  cos  B  cos  c 
cot  b  sin  a  =  cot  B  sin  C  +  cos  C  cos  a 
cot  6  sin  c  =  cot  B  sin  A  +  cos  A  cos  c 
cot  c  sin  a  =  cot  C  sin  B  -|-  cos  B  cos  a 
cot  c  sin  6  =  cot  C  sin  A  +  cos  A  cos  b 


(2) 
(•^) 
(4) 
(r») 
(6) 


i 


9^ 


I. 


ri* 


280 


SPHERICAL   TRIGONOMETRY. 


EXAMPLES. 


1.  If  a,  6,  c  be  the  sides  of  a  spherical  triangle,  a',  &',  c' 
the  sides  of  its  polar  triangle,  prove 

sin  a :  sin  6  :  sin  c  s=  sin  a' :  sin  &' :  sin  c'. 

2.  If  the  bisector  AD  of  the  angle  A  of  a  spherical 
triangle  divide  the  side  BC  into  the  segments  CD  =  6', 
BD  =  c',  prove 

sin  &  :  sin  c  =  sin  6' :  sin  c'. 

3.  If  D  be  any  point  of  the  side  BC,  prove  that 

cot  AB  sin  DAC  +  cot  AC  sin  DAB  =  cot  AD  sin  BAC. 
cot  ABC  sin  DC  +  cot  ACB  sin  BD  =  cot  ADB  sin  BC. 

4.  If  a,  /3,  y  be  the  perpendiculars  of  a  triangle,  prove  that 

sin  a  sin  a  =  sin  b  sin  /?  =  sin  c  sin  y. 

5.  In  Ex.  4  prove  that 


sin  a  cos  a  =  Vcos'*  b  -\-  cos^  c  —  2  cos  a  cos  b  cos  c. 

194.  Useful  FonnulfiB.  —  Several  other  groups  of  useful 
formulae  are  easily  obtained  from  those  of  Art.  191 ;  the 
following  are  left  as  exercises  for  the  student : 


sin  a  cos  B  =  cos  b  sin  c  —  sin  b  cos  c  cos  A 
sin  a  cos  C  =  sin  &  cos  c  —  cos  b  sin  c  cos  A 
sin  b  cos  A  =  cos  a  sin  c  —  sin  a  cos  c  cos  B 
sin  b  cos  C  =  sin  a  cos  c  —  cos  a  sin  c  cos  B 
sin  c  cos  A  =  cos  a  sin  6  —  sin  a  cos  b  cos  C 
sin  c  cos  B  =  sin  a  cos  b  —  cos  a  sin  b  cos  C 


(1) 

(2) 

i^) 
(4) 
(5) 
(6) 


FORMULA  FOR   THE  HALF  ANGLES. 


281 


.t  /.» 


(1) 
(2) 

{^) 
(4) 
(5) 
(6) 


Applying  these  six  formulae  to  the  polar  triangle,  we 
obtain  the  following  six : 


sin  A  cos  b  =  cos  B  sin  C  +  sin  B  cos  C  cos  a 
sin  A  cos  c  =  sin  B  cos  C  +  cos  B  sin  C  cos  a 
sin  B  cos  a  =  cos  A  sin  C  4-  sin  A  cos  C  cos  b 
sin  B  cos  c  =  sin  A  cos  C  +  cos  A  sin  C  cos  b 
sin  C  cos  a  =  cos  A  sin  B  4-  sin  A  cos  B  cos  c 
sin  C  cos  b  =  sin  A  cos  B  +  cos  A  sin  B  cos  c 


(7) 

(8) 

(9) 
(10) 

(11) 
(12) 


195.  FormulsB  for  the  Half  Angles.  —  To  express  the 
sine,  cosine,  and  tangent  of  half  an  angle  of  a  spherical 
triangle  in  terms  of  the  sides. 

I.  By  (4)  of  Art.  191  we  have 

.         cos  a  —  cos  &  cos  C        ^         o    •    2^.    /  A    4.    An\ 

COS  A  = — =  1  —  2  sin^  -  (Art.  49) 

sin  b  sin  c  2 

o   •  2  A      ^      COS  a  —  COS  b  cos  c 

2  Sin  &  sine  v,;', 

cos  (6  —  c)  —  cos  a 
sm  0  sin  c 


sin 


2 A  =  sin^(a+&-c)sin|(a-&  +  c)  ^^^^  ^^^ 


2 


sin  b  sin  c 


Let  2s  =  a  +  64-c;  so  that  s  is  half  the  sum  of  the  sides 
of  the  triangle ;  then 

a +  b —  c  =  2{s  —  c),  and  a  —  b-{-c  =  2{s  —  b). 

•  2  A      sin  (s  —  &)  sin  (s  —  c) 

2  sin  &  sin  c  ^ 


.-.    sin  ^  =x^^"i''  ~  ^^  ^^"  il".'^) 
2      \  sin  b  sin  c 


(1) 


m\  \ 


282 


SPHERICAL    TUIGONOMETRY. 


11  V    1 


Advancing  letters, 


sm 


—  c)  sin  (s  —  g) 


B  _    /sin  (.s  —  c)  si 
2       >/  sin  c  si 


sma 


.     .     .     (2) 


II. 


.    C        /sin  (.s  — a)  sin  (s  — 6)  ,o\ 

sin  — =\/ \__— ^_-— ^ L  ...     (3) 

2      \  sin  a  sin  6 

J 

2  CDs'*  —  =  1  -f-  cos  A (Art.  49) 


=  1  + 


COS  a  —  cos  h  cos  c 
sin&  sine 


,*.  cos 


__  cos  a  —  cos  (6  4-  c) 
sin  h  sin  c 

2A_  sin  \{a  +  64-0)  sin  ^(6  4-  c  —  a) 
sin  6  sin  c 

_  sin s  sin(g  —  ») 
sin 6  sine 


cos 


-^ 


sin  3  sin  (8  —  a) 


sin  h  sin  e 


(4) 


Advancing  letters, 


B        /sinssin(s  — ^) 

cos  —  =\/ : ^ '- 

2      \      sm  e  sin  a 


•         • 


(5) 


cos 


C  _    Isin  s  sin  (&•  —  e)  ,n.^ 

2      Ai      sin  a  sin  6 

III.   By  division,  we  obtain 
tan 


tan 


A  _    /sin {h  —  b)  sin (s  —  c) 
2       >/      sin  s  sin  (.s  —  a) 

]^  —    /sill  ('**  —  c)  sin  (■*»  —  «) 
2      Al       sins  sin (s  —  i^) 


(7) 

(8) 


tang=J'^^:^'^"^'~^   ...     (9) 
2      \       sinssinf.s  — c)  ^ 


.     (4) 


.     (5) 
.     (6) 


.  (7) 
.  (8) 
•     (9) 


FORMULA      OR   THE  HALF  ANGLES.  283 

Sch.  The  positive  sign  jnusfc  be  given  to  the  radicals  in 
each  case  in  this  aj'ticle,  because  |-A,  ^B,  ^C  are  each  less 
than  90°. 

Cor.l.  tan^tan?  =  ^^^-^^    ....     .(10) 

2         2  sins  ^     ^ 

tan  — tan     = — >^ ^ (11) 

2        2  sins  ^     ^ 

.      C.      A      sin(.s'  — ?/)  ,^fv, 

tan  -  tan— =  — ^ '- (12) 

2        2  sins  ^     ' 

A       A 

Cor.  2.    Since  sin  A  =  2  sin  —  cos  — , 

.     A         2Vsins  sin(s  —  «)  sin(s  — />)  sin(s  —  c)  .^o\ 

.-.  sin  A    = —-^ > i — r^^ ^ >^ ^      ,   (1,>) 

sni6  sine 
-      2w  .... 

— •  •••••••••«»l   ATI:  J 

Sin  6  sin  c 
where  «^  =  sinssin(s  — a)  sin(s  — 6)  siii(s  —  c). 

EXAMPLES. 

M     -n  •  2A      1  — cos^a— ('os-6— cos^c4-2cosacos&cosc 

1.  Prove  siirA= — r-^ — r— ^ - 

snrb sure 

_        4ri^ 
sin^&  sin^c 
where  4  w*  =  1  —  cos^  a  —  cos^  6  — j3os^  c  +  2  cos  a  cos  b  cos  c. 

C  C 

2.  Prove  cose  =  cos(a+ &)sin^ — h  cos  (a  —  &)  cos^*  -  • 

o    T3            .    A   .    B    .    C     sin(s— a)  sin(s  — &)sin(s  — c) 
o.    Prove  sin  —  sin—  sin  -  =  — -^^ ^ -^^ ^ ^ -. 

2        2        2  sin  a  sin  6  sm  e 

4.  Prove   cos  ^  4-cosB  ^sin  (» +  &). 

1  — cosC  sine 

e    T>  ^cosA4  cosB   •    /        ,x    .  A 

5.  Prove  2 sin  (a  —  b)  sine  =  0. 

1  —  cui  U  ^^*v 

^    -o  cos  A  —  cos  B      sin  (a  ~  b) 

6.  Prove   r^ ^ -' 

1  +  cosC  sine 


II 


vm? 


284  SPHERICAL   TRIGONOMETRY. 

196.  FormulflB  for  the  Half  Sides.  —  To  express  the  sine, 
cosine,  and  tangent  of  half  a  side  of  a  spherical  triangle  in 
terms  of  the  angles. 

By  (1)  of  Art.  192,  we  have 

c^sA  +  cosBcosC^^_23i^.a         (Art.  49) 
smBsiiiC  2         ^  ^ 

.-.  2  3111=^^=      cosA4-cos(B  +  C) 
2  sin  B  sill  C 

.    sin2^  =  - cosi(A  +  B+C)cos^(B+C-A)      .^      .^. 
2  sin  B  sin  C  ^       *      ^ 

Let  2S  =  A  +  B  +  C;  then  B  +  C  -  A  =  2  (S  -  A). 

Proceeding  in  the  same  way  as  in  Art.  195,  we  find  the 
following  expressions  for  the  sides,  in  terras  of  the  three 
angles : 

sin«  =  -.CcosSc^3S^A) .^. 

2      \  sin  B  sin  0  ^ 

.    b         I     cos  S  cos  (S  —  B)  ,o\ 

sin-  =  ^/ ; — —\ — - — ^ (2) 

2      A/  sinCsinA  ^  ^ 

2      \  sinAsinB  ^  ^ 

cos  -  =  ^  /co«(S-B)cos(S-C)  ... 

2      \  sinBsinC  .....     v  / 

2      \  smCsinA  .....     vu; 

cos^  =  J^^^S:^S«pISEl} (6) 

2      \  SinAsinB  ^  ^ 

— 

tan  ^  =  J-  -^^'^'^  S  ^os  ( S  -  Al_      ....     (7) 
2      \      cos  (S  -  B)  cos  (S  -  C)  '^ 


Itill'l! 


FOHMUL^  FOR    THE  HALF  SIDES. 


286 


tan 


H- 


cQsS  cos(S -r  B) 
co8(S--C)cos(S-A) 


...    .     (8) 


.  (1) 

.  (2) 

.  (3) 

.  (4) 

■  (5) 

.  (6) 

.  (7) 


tan£  =  J cos  S  cos  (S  -  C) 

2      \     cos  (S- A)  cos  (S- 


(S  -  A)  cos  (S  -  B) 


•        •        •         •  V      / 


Sch.  1.  These  formulae  may  also  be  obtained  immediately 
from  those  of  Art.  195  by  means  of  the  polar  triangle. 

Sch.  2.  The  positive  sign  must  be  given  to  the  above 
radicals,  because  -,  -,  -,  are  each  less  than  90°. 

Sch.  3.  These  values  of  the  sines,  cosines,  and  tangents 
of  the  half  sides  are  always  real. 

For  S  is  >  90°  and  <  270°  (Art.  184),  so  that  cos  S  is 
always  negative. 

Also,  in  the  polar  triangle,  any  side  is  less  than  the  sum 
of  the  other  two  (Art.  184). 

.-.     TT  —  A<7r  —  B+TT—  C. 

.-.   B  +  C-A<7r. 

.'.  cos  (S  —  A)  is  positive. 

Similarly,    cos  (S  —  B)  and  cos  (S  —  C)  are  positive. 


Cor.   Since  sin  a  =  2  sin  -  cos  % 


smct: 


.2V-cosScos(S-A)cos(S-B)cos(S-C) 


sin  B  sin  C 


(10) 


2N 


sin  B  sin  C 


where  N=  V—  cos  S  cos  (S  --  A)  cos  (S  —  B)  cos  (S  —  C). 


m 

''-'^B- 

286 


SPHERICAL   TRIGONOMETRY. 


EXAMPLBS. 


1.   Prove  cosC=— cos(A  +  B)cos^^ — cos(A— B)  sin^-- 


a 


b  .:.    C 


2.   Prove  sin  -  sin  -  sin    =  — 

9.  O.  9        si 


—  N  cos  S 


2      sin  A  sin  B  sin  C 


where  N  =  V—  cos  S  cos  (S  —  A)  cos  (S  —  B)  cos  (S  —  C). 


197.  Napier's  Analogies. 
Let  m 


or 


sin  A  _  sinB 
sin  a       sin  b 

sin  A  +  sinB 
sin  a  +  sin  6 

sin  A  —  sin  B 


(Art.  190)  (1) 
(Algebra)  (2) 
...     (3) 


sin  a  —  sin  b 

cos  A  +  cos  B  cos  C  =  sin  B  sin  C  cos  a   (Art.  192) 

=  m  sin  C  sin  b  cos  a,  by  (1 )  (4) 

and  cos  B  +  cos  C  cos  A  =  sin  C  sin  A  cos  b 

=  m  sin  C  sin  a  cos  &  .     .     (5) 
.-.  (cos  A +  cosB)(l  4-cosC)  =  msinC  sin  (a  +  6),  (6) 


Dividing  (2)  by  (6), 

sin  A  4-  sin  B 


from  (4)  and  (5) 


sin  a  4-  sin  6    1+  cos  C 


cos  A  +  cos  B       sin  (a  +  b)         sin  C 


.-.  tani(A-j-B)  = 


cosi(a-6)^^^C 
cos  ^  (a  +  6)        2 
(Arts.  46,  46,  and  49) 


(7) 


Similarly,       tan^  (A  -  B)  =  -"^^^i  (^  -  ^)  cot^  .     .     (8) 
^'  ^^  '     sini(a  +  &)       2  ^  ' 


V 


DELAMBRE'8  ANALOGIES. 


287 


I  ♦ 


-C). 

30)  (1) 
ra)  (2) 

.     (3) 

rt.  192) 

Kl)(4) 

.     (5) 

(6) 
(5) 

osC 

_— -I— ■■  • 

0 


(7) 


49) 


Writing  71  —  A  for  a,  etc.,  by  Art.  184,  we  obtain  from 
(7)  and  (8) 

ScL  The  formulae  (7),  (8),  (9),  (10)  are  known  as 
Napier's  Analogies,  after  their  discoverer.  The  last  two 
may  be  proved  without  the  polar  triangle  by  starting  with 
the  formulae  of  Art.  191. 

Cor.  Ill  any  spherical  triangle  ivhose  parts  are  positive, 
and  less  than  180°,  the  half-sum  of  any  two  sides  and  the  half- 
sum  of  their  opposite  angles  are  of  the  same  species. 

C 
For,  since  cos  ^  (a  —  6)  and  cot  —  are  necessarily  positive, 

therefore  by  (7)  tan^(A4-B)  and  cos  ^  (a +  6)  are  both 
positive  or  both  negative. 

.-.  |(A  +  B)  and  ^ (a  +  6)  are  both  >  or  both  <  or  both 
=  90°. 

198.  Delambre's  (or  Gauss's)  Analogies. 
sini(A4-B) 

A       B  ,         A   .    B 

=  sm  2  ^^^  2  +  ^°^  2  ^^^  2 


Vsin  {s  —  b)  sin  (.s  —  c)     /sin  s  sin  (s  —  6) 
sin  6  sin  c  ^      sin  c  sin  a 


+ 


.     (8) 


/sin8sin(.t-a)     /sin  (6--c)  sin  (s  -  ct)    ,^^^^  jg^^ 
\       sin  6  sin  c       Af  sin  c  sin  a 

_  sin  (s  —  &)  +  sin  {s  —  a)     /sin  s  sin  {s  —  c) 
sin  c  A(      sin  a  sin  b 

^cos^{a-b)G (Art..  45  and  195) 


•  1'' 


cos 


2 


k 


288 


SPHERICAL    TEW ONOMETR  Y. 


.:  sin^  (A  +  ]i)  cos- —  oos^  (d— 6)  cos  -     .     .     .     (1) 

Similarly,  we  obtain  the  following  three  equations  : 

c  C 

sin^  (A  —  B)  sin-  =  sin^  (a  —  6)co8  -     ...     (2) 

c  C 

cos  |(  A  4-  B)  cos  -  =  cos  I  (a  +  6)  sin  -     .     .     .     (3) 


c        •  C 

cos  i  (A  —  B)  sin  -  =  sin  ^  (a  +  6)  sin  — 


(4) 


8c\i.  1.  When  the  sides  and  angles  are  all  less  than  180", 
both  members  of  these  equations  are  positive. 

Sell.  2.  Napier's  analogies  may  be  obtained  from  De- 
lambre's  by  division. 

Note.  — Delarabre's  analogiea  were  discovered  by  him  in  1807,  and  publiBhed  in 
the  Connaissance  des  Temps  for  1809,  p.  44:J.  They  were  subsequently  discovered 
independently  by  Gauss,  and  published  by  bim,  and  arc  sometimes  improperly 
called  Gauias's  equutious.  Both  systems  may  be  proved  geometrically.  The 
geometric  proof  is  the  one  originally  given  by  Delambre.  It  was  rediscovered  by 
Professor  Crofton  in  1869,  and  published  in  the  Proceedings  of  the  Loudon  Mathe- 
matical Society,  Vol.  HI.  [Casey '■  Trigonometry,  p.  41], 


EXAMPLES. 

In  the  right  triangle  ABC,  in  which  C  is  the  right  angle, 
prove  the  following  relations  in  Exs.  1-45  : 

1.  sin^acos^&  =  sin  (c  +  &)  sin  (c  — &). 

2.  tan*  a  :  tan*  6  =  sin*c  —  sin*&  :  sin*c  —  sin*  a. 

3.  cos'  a  cos*  B  =  sin*  A  —  sin*  a. 

4.  cos*  A  +  cos*  c  =  cos*  A  cos*  c  +  cos*  a. 

5.  sin*  A  — cos*  B  =  sin*  a  sin*  B. 

6.  If  one  of  the  sides  of  a  right  triangle  be  equal  to  the 
opposite  angle,  the  remaining  parts  are  each  equal  to  90°. 


EXAMPLES. 


289 


.     (1) 

.     (2) 

.     (3) 

.     (4) 
Ml  180°, 


om 


De- 


mbllshed  in 
discovered 
improperly 
:ally.  The 
Bcovered  'oy 
idon  Matbe- 


;ht  angle, 


7.    Tf  the  angle  A  of  a  right  triangle  be  acute,  show  that 
the  difference  of  the  sides  which  contain  it  is  less  than  90°. 


8.   Prove 


.      B      sin  (s  —  a) 

tan     = \ ^« 

2  sins 


9.    Prove  (1)  2n  =  sinasin&;   (2)  2N  =  sinasinB. 
10.   Prove     sin^  a  sin^  b  =  sin*  a  +  sin^  b  —  sin'  c. 


11.  Prove 

12.  Prove 


^^^,A^sin(c-6)^ 

2      sin(c  +  6)       y  •   •    ' 

2  sin2^  =  sin2^(a  +  b)  +  sinH(a  -  b) . 

4W 


13.  In  a  spherical  triangle,  if  c  =  90°,  prove  that 

tan  a  tan  b  +  sec  C  =  0. 

14.  In  a  spherical  triangle,  if  c  =  90°,  prove  that 

sin^p  =  cot  ^  cot  <^,  , 

where  p  is  the  perpendicular  on  c,  and  0  and  <^  are  the  seg- 
ments of  the  vertical  angle.  ,;:      , 

15.  Show  that  the  ratio  of  the  cosines  of  the  segments 
of  the  base  made  by  the  perpendicular  from  the  vertex  is 
equal  to  the  ratio  of  the  cosines  of  the  sides. 

16.  If  B  be  the  bisector  of  the  hypotenuse,  show  that  : 

sin^'a.  4-  sink's  ':      v' 


sin'B  = 


4  cos 


2C 


Lial  to  the 
L  to  90°. 


17.   Prove 


tanS  =  cot- cot-. 


18.  Construct  a  triangle,  being  given  the  hypotenuse  and 
(1)  the  sum  of  the  base  angles,  and  (2)  the  difference  of 
the  base  angles. 

19.  Given  the  hypotenuse  and  the  sum  or  difference  of 
the  sides :  construct  the  triangle. 


290 


SPHEliICA L    TlilGONOMETU  Y. 


20.   Given  the  sum  of  the  sides  a  and  h,  and  the  sum 
of  the  base  angles :  solve  the  triangle. 


oi     c!u       i.u  1.    •    A     Vsinc-f  sm  u  +  Vsinc  — sin  M 
21.   Shovir  that  sin  — =  -- ^ 

9 


2  Vsin  c 


22.   sin 


'  \2C08/ 


h  sin  c 


23.   cos 


V 2 cos 6  sine 


24.    sin  (a  +  6)  tan  ^  ( A.  +  H)  =  sin  (a  —  6)  cot  ^ ^  -  B). 

cos  a  -\-  cos  h 


2o.   sin(A  +  B): 
26.    sin(A-B) 


27.  cos(A  +  l.)  =  - 

28.  cos(A-B)  =  — 


1  +  cos  a  cos  h 

cos  ft  —  cos  a 
1  —  COS  a  cos  6 

sin  a  sin  6 


1  4-  cos  a  cos  6 
sin  rt  sin  6 


cos  a  cos  ft 


29.    sin2  '^  =  sin^  ^  cos''  -  +  cos"  ^  sin''  -. 
2  2         2  2        2 


30.   sin  (c  -  6)  =  sin  (c  +  ft)  tan*  _. 


A 


B 


31.    sin  (a  —  ft)  =  sin  a  tan sin  ft  tan 

^  ^  2  2 


32.    sin  (c  —  a)  =  cos  a  sin  6  tan 


B 


33.  If  ABC  is  a  spherical  triangle,  right-angled  at  C, 
and  cos  A  =  cos"  a,  show  that  if  A  be  not  a  right  angle, 
ft-fc  =  ^7r  or  fTT,  according  as   ft   and   c  are   both   <   or 

both  >  |. 


EXAMPLES. 


291 


B). 


34.  If  «,  /8  be  the  arcs  drawn  from  the  riyht  angle 
respectively  perpendicular  to  and  bisecting  the  hypotenuse 
c,  show  that 

8in=*-^(l  +  sin2«)  =  sin»/?. 

35.  In  a  triangle,  if  C  be  a  right  angle  and  D  the  middle 
point  of  AB,  show  that 

c 
4  cos''- sin^  CD  =  sin'*  a  +  sin*  6. 

In  a  right  triangle,  if  p  be  the  length  of  the  arc  drawn 
from  the  right  angle  C  perpendicular  to  the  hypotenuse 
AB,  prove : 

36.  cot*^7  =  cot- a  4- cot^  6. 

37.  cos^i)  =  cos^  A  +  cos'^B. 

38.  tan- a  =  ^  tan  a'  tan  c. 

39.  tan'*  b  =  ±  tan  b'  tan  c. 

40.  tan'* a :  tan'* 6  =  tana' :  tan  b'. 

41.  sin^/)  =  sin  a'sin  6'. 

42.  sin/)  sin  c=  sin  a  sin  6. 

43.  tan  a  tan  b  =  tan  c  sin  p. 

44.  tan'*  a  +  tan  ^6  =  tan^c  cos'^p. 

45.  cot  A  :  cot  B  =  sin  a' :  sin  b'. 

In  the  oblique  triangle  ABC,  prove  the  following : 

46.  If  the  difference  between  any  two  angles  of  a  tri- 
angle is  90°,  the  remaining  angle  is  less  than  90°. 

47.  If  a  triangle  is  equilateral  or  isosceles,  its  polar  tri- 
angle is  equilateral  or  isosceles. 

48.  If  the  sides  of  a  triangle  are  each  -,  find  the  sides 
of  the  polar  triangle. 


41 


^92 


SPHERICAL   TRIGONOMETRY. 


40.   If  in  a  triangle  the  side  a  =  90°,  show  that 
cos  A  4-  cos  B  cos  C  =  0. 

50.  If  0  and  6*  are  the  angles  which  the  internal  and  ex- 
ternal bisectors  of  the  vertical  angle  of  a  triangle  make  with 
the  base,  show  that 


cosO  = 


cos  A  ~  cos  B 


2  cos 


C 


■J  and  cos  6'  = 


cos  A  +  cos  B 
2sin^ 


oos  A. 

51.  Given  the  base  c  and  - — —  =  —  cos  C :  find  the  locus 

of  the  vertex. 

52.  Prove  4N^  =  1  —  cos'^A  —  cos^B  —  cos^C 

—  2  cos  A  cos  B  cos  C. 

53.  If  p,  q,  r  be  the  perpendiculars  from  the  vertices  on 
the  opposite  sides,  show  that 

(1)  sin  a  sin  J)  =  sin  6  sin  7  =  sin  c  sin  r  =  2n. 

(2)  sinAsinj9  =  sinB  sin  7  =  sinCsinr  =  2N. 

54.  Prove  8  n''  =  sin^  a  sin^  b  sin^  c  sin  A  sin  B  sin  C. 
sin'^  A  +  sin'^  B  -f-  sin*^  C      1  +  cos  A  cos  B  00s  C 


55.   Prove 


sin^  a  -|-  sin^  6  +  sin^  c         1  —  cos  a  cos  &  cos  c 


56.   If  I  be  the  length  of  the  arc  joining  the  middle  point 
of  the  base  to  the  vertex,  find  an  expression  for  its  length 


in  terms  of  the  sides. 


Ans.  cosh 


cos  (I  -f-  cos  h 


2  cos 


2 


57.    If  CD,  CD'  are  the  internal  and  external  bisectors  of 
the  angle  C  of  a  triangle,  prove  that 

^  /-,T^      cx)t  a  "f-  cot  b        -      ^  / ,  T^.      cot  a  ~  cot  & 
cot  CD  = — ,  and  cot  CD'  = 


2cos^ 
2 


2sin^ 


EXAMPLES. 


298 


58.  Show  that  the  angles  $  and  6',  made  by  the  bisectors 
of  the  angle  C  in  Ex,  55  with  the  opposite  side  c,  are  thus 
given : 

cota  — cot6  .    ^-p,  _ 

..  cot^  = — sin  CD,  ! 


2  sin 


C 


.,f.  .,■ 


cotrt  +  cotft   . 
cot  B  = —  sin  CD'. 


cos- 


C 


59.    Show  that  the  arc  intercepted  on  the  base  by  the 
bisectors  in  Ex.  55  is  thus  given : 


cotDD' 

60.    Prove  that 

cos^6  —  cos^c 


sin'^A  —  sin^B 
2  sin  A  sinB  siiiC 


cos^c  —  cos^a 


cos  6  cot  B  —  cos  c  cot  C   cos  c  cot  C  —  cos  a  cot  A 

cos*  a  —  cos*  6 


cos  a  cot  A  —  cos  h  cos  B 


61.  If  s  and  s'  are  the  segments  of  the  base  made  by  the 
perpendicular  from  the  vertex,  and  m  and  m'  those  made 
bj-  the  bisector  of  the  vertical  angle,  show  that 


tan- 


.S'' 


m  —  )/ 


.,a  —  h 


tan =  tair 

2  2  2 


62.  Prove 

sin  6  sin  c  +  cos  h  cos  c  cos  A  =  sin  B  sin  C  —  cos  B  cos  C  cos  a. 

63.  Show  that  the  arc  I  joining  the  middle  points  of  the 
two  sides  a  and  6  of  a  triangle  is  thus  given : 


cosZ  = 


1  +  cos  a  4-  cos  h  ■\-  cos  c 

■  ■  —  ■  -  » 

4  cos    cos- 
2       2 


SPHERIC  A  L   TRIGONOMETR  Y. 

64.  If  the  side  c  of  a  triangle  be  90°,  and  S  the  arc  drawn 
at  right  angles  to  it  from  the  opposite  vertex,  show  that 

COt^SrzzCOt^A  +  COt^B. 

65.  Prove  that  the  angle  (f>  between  the  perpendicular 
from  the  vertex  on  the  base  and  the  bisector  of  the  vertical 
angle  is  thus  given  : 

tan  0  =  «^«  ii'^^^^ltan  ^  ( A  -  B). 
^     cos  I  (a +  6) 

66.  In  an  isosceles  triangle,  if  each  of  the  base  angles  be 
double  the  vertical  angle,  prove  that 

cosacos-  =  cosf  c  + -V 

67.  If  a  side  c  of  a  triangle  be  90°,  show  that 

(1)  cotacot&  +  cos  C  =  0. 

(2)  cos  S  cos  (S  -  C)  +  cos  (S  -  A)  cos  (S  -  B)  =  0. 

68.  In  any  triangle  prove 

cos  a  —  cos  b     sin  (A  —  B)  _  ^ 
1  —  cos  c  sin  C 

69.  tan|(A+B):tan|(A-B) 

=  tan  ^  (a  +  6)  :  tan  ^  (a  —  6). 

70.  tan  ^  (A  -f-  a)  :  tan  |(  A  —  a) 

=  tan |(B  +  6)  :  tan ^ (B  - &). 

71.  If  the  bisector  of  the  exterior  angle,  formed  by  pro- 
ducing BA  through  A,  meet  the  base  BC  in  D',  and  if  BD 
=  c",  CD'  =  b",  prove 

sin  ft  :  sin  c  =  sin  ft" :  sin  c". 

72.  If  D  be  any  point  in  the  side  BC  of  a  triangle,  prove 

sinBI)_sinBAD    sinC 
sin  CD     sin  cad"  sin  b' 


EXAMPLES. 


295 


0. 


73.  If  A  =  a,  show  that  B  and  b  are  either  equal  or 
supplemental,  as  also  C  and  c. 

74.  If  A  =  B  +  C,  and  D  be  the  middle  point  of  a,  show 
that  a  =  2  AD. 

75.  When   does   the   polar   triangle   coincide   with    the 
primitive  triangle? 

76.  If  D  be  the  middle  point  of  c,  show  that 

cos  a  -f-  COS  6  =  2  cos     cos  (^I). 

2  \,,,^  ■, 

77.  In  an  equilateral  triangle  show  that 

(1)  2  cos- sin  — =  1. 
^  ^  2        2 

(2)  tan2^  +  2cosA=l. 

78.  If  &  4-  c  =  TT,  show  that  sin  2  B  +  sin  2  C  =  0. 

79.  Show  that 

sin  b  sin  c  +  cos  &  cos  c  cos  A  =  sin  B  sin  C  —  cos  B  cos  C  cos  a. 

80.  If  T>  be  any  point  in  tlie  side  BC  of  a  triangle,  show 

that 

cos  AD  sin  a  =  cos  c  sin  DC  +  cos  b  sin  BD. 

81.  Prove  cos^   ==cos''|^(a+6)sin''    -|-t^os^|(a— 6)cos*— • 

82.  ''      sin2^=sin4(a+&;sin2^+sin2^(a-6)cos2p. 

a  Z  ^ 

83.  "      sin  s  sin  (s  —  a)  sin  (s  —  b')  sin  (s  —  c) 

=  ^(1  —  cos^a  —  cos^6  —  cos^c  4- 2  cos  a  cos  &  cose). 

84.  If  AD  be  the  bisec^'^r  of  the  angle  A,  prove  that 

(1)  cos  B  4-  cos  ^    ^     sin  -  sin  ADB  cos  AD. 


(2)  cos  C  -  cos  B  =  2  cos  ^  cos  ADB. 


296 


SPHERICAL   TRIGONOMETRY. 


85.  Prove  cos  a  sin  6  =  sin  a  cos  6  cos  C  +  cos  A  sin  c. 

86.  "       sin  C  cos  a  =  cos  A  sin  B  +  sin  A  cos  B  cos  C. 


87.   In  a  triangle  if  A  =  ^,   B  =  ^,   C  =  ^,   show  that 

O  ij  M 


ir 


a  +  6  -|-c=  - 


88.   Prove  sin(S-A): 


1  +  cos  a  —  cos  b  —  cos  c 

.        a   •    b   ■    c 
4  cos  -  sin  -  sin  - 


89.  If  8  be  the  length  of  the  arc  from  the  vertex  of  an 
isosceles  triangle,  dividing  the  base  into  segments  a  and  /3, 
prove  that 

tan  -  tan  "  =  tan         -  tan  —^ — 


90.  If  b  =  c,  show  that 

.    a  A 

sin  -  cos  — 

2  2 

sin  b  = r»  and  sin  B  = 

.    A  a 

sin  —  cos  - 

2  2 

91.  If  AB,  AC  be  produced  to  B',  C,  so  that  BB',  CC 
shall  be  the  semi-supplements  of  AB,  AC  respectively, 
prove  that  the  arc  B'C  will  subtend  an  angle  at  the  centre 
of  the  sphere  equal  to  the  angle  between  the  chords  of 
AB,  AC. 


PRELIMINAHY  OBSERVATIONS. 


297 


■.)  f. 


CHAPTER   XI. 


SOLUTION  OF  SPHERIOAL  TEIANGLES. 


/   ;■•'.. 


199.  Preliminary  Observations.  —  In  every  spherical  tri- 
angle there  are  six  elements,  the  three  sides  and  the  three 
angles,  besides  the  radius  of  the  sphere,  which  is  supposed 
constant.  The  solution  of  spherical  triangles  is  the  process 
by  which,  when  the  values  of  any  three  elements  are  given, 
we  calculate  the  values  of  the  remaining  three  (Art.  184, 
Note). 

In  making  the  calculations,  attention  must  be  paid  to  the 
algebraic  signs  of  the  functions.  When  angles  greater  than 
90°  occur  in  calculation,  we  replace  them  by  their  supple- 
ments ;  and  if  the  functions  of  such  angles  be  either  cosine^ 
tangent,  cotangent,  or  secant,  we  take  account  of  the  change 
of  sign. 

It  is  necessary  to  avoid  the  calculation  of  very  small 
angles  by  their  cosines,  or  of  angles  near  90°  by  their  sines, 
for  their  tabular  differences  vary  too  slowly  (Art.  81).  It 
is  better  to  determine  such  angles,  for  example,  by  means 
of  their  tangents. 

We  shall  begin  with  the  right  triangle;  here  two  ele- 
ments, in  addition  to  the  right  angle,  will  be  supposed 
known. 


"'U 


SOLUTION  OF  RIGHT  SPHERICAL  TRIANGLES. 

200.  The  Solution  of  Right  Spherical  Triangles  presents 
Six  Cases,  which  may  be  solved  by  the  formulae  of  Art. 
185.  If  the  formula  required  for  any  case  be  not  remem- 
bered, it  is  always  easy  to  find  it  by  Napier's  Kules  (Art, 


298 


SPHERICAL    TRIGONOMETRY. 


186).     In  applying  these  rules,  we  must  choose  the  middle 
part  as  follows : 

When  the  three  parts  considered  are  all  adjacent,  the 
one  between  is,  of  course,  the  middle  part.  When  only 
two  are  adjacent,  the  other  one  is  the  middle  part. 

Let  ABC  be  a  spherical  triangle, 
right-angled  at  C,  and  let  a,  b,  c 
denote  the  sides  opposite  the  angles 
A,  B,  vJ,  respectively. 

W(^  shall  assume  that  the  parts 
are  aU  positive  and  less  than  180° 
(Art.  182). 

201.  Case  I.  —  Given  the  hypotenuse  c  and  an  angle  A ;  to 
find  a,  b,  B. 

By  (•^).  (5),  and  (8)  of  Art.  185,  or  by  Napier's  Rules, 
we  have 

sin  a  =  sin  c  sin  A, 

tan  b  =  tan  c  cos  A, 

cot  B  =  cos  c  tan  A. 

Since  a  is  found  by  its  sine,  it  would  be  ambiguous,  but 
the  ambiguity  is  removed  because  a  and  A  are  of  the  same 
species  [Art.  187,  (1)].  B  and  b  are  determined  imme- 
diately without  ambiguity. 

If  a  be  very  near  90°,  w^e  commence  by  calculating  the 
values  of  b  and  B,  and  then  determine  a  by  either  of  the 
formulae 

tan  a  =  sin  b  tan  A,  tan  a  =  tan  c  cos  B. 

Check.  —  As  a  final  step,  in  order  to  guard  a<r  mst  numer- 
ical errors,  it  is  often  expedient  to  check  the  logarithaiic 
work,  which  may  be  done  in  every  case  without  the  neces- 
sity of  new  logarithms.  To  check  the  work,  we  make  up 
a  formula  between  the  three  required  parts,  and  see  whether 


» 


CASE  I.I 


299 


B 


h 


it  is  satisfied  by  the  results.     In  the  present  case,  when  the 
three  parts  a,  h,  B  have  been  found,  the  check  formula  is 

sin  a  =  tan  &  cot  B   ....     [(6)  of  Art.  185] 
Ex.  1.    Given  c  =  81°  29'  32",  A  =  32°  18'  17"  ;  find  a,  h,  B. 


Solution. 


log  sine    =9.9951945 
log  sin  A  =  9.7278843 

log  sin  a  =  9.7230788 

.-.  a  =31°  54' 25". 

log  cose    =9.1700960 
log  tan  A  =  9.8009157 


log  cot  B  =8.9710117 
.-.  B  =  84°  39'  21".33. 


log  tan  e    =10.8250982 
log  cos  A  =    9.9269687 

log  tan  6   =10.7520669 

.-.  b   =  79°  51' 48".65. 

Check. 

log  tan  6   =10.7520669 
log  cot  B  =    8.9710117 

log  sin  a  =    9.7230786 


Ex.  2.    Given  e  =  110°  46'  20",  A  =  80°  10'  30" ;  find  a,  &,  B. 
Ans.  a  =  67°  5'  52".7,  b  =  155°  40'  42".7,  B  =  153°  58'  24".5. 

202.   Case  II.  —  Oiven  the  hypotenuse  e  and  a  side  a  ;  to 
find  b,  A,  B. 

By  (1),  (3),  (4)  of  Art.  185,  or  by  Napier's  Rules,  we 
have 

tana 

COSI5  = 

sin  c 


,      cos  c     .     A       Sin  a         n 
COS  0  = ,  sm  A  =  — — ,  cos  B  = 


cos  a 


tan  c 


The  check  formula  involves  6,  A,  B ;  therefore,  from  (9) 
of  Art.  185  we  have 

cos  B  =  sin  A  cos  b. 

In  this  case  there  is  an  apparent  ambiguity  in  the  value 
of  A,  but  this  is  removed  by  considering  that  A  and  a  are 
always  of  the  same  species  (Art.  187). 


300 


SPHERICAL   TRIGONOMETRY. 


Ex.  1.   Gi  ven  c  =  140°,  a  =  20° ;  find  6,  A,  B. 

Sohition. 


log  cose     =9.8842540- 
colog  cos  a  =  0.0270142 

log  cos  6     =9.9112682- 

.-.  h     =  144°  36' 28".4. 

log  tan  a     =  9.5610659 
colog  tan  c  =  0.0761865- 

logcosB    =9.6372524- 

.-.  B    =  115°  42' 23".8. 


log  sin  a    =  9.5340517 
colog  sine  =0.1919325 

log  sin  A    =9.7259842 

.-.  A    =32°8'48".l. 

Check. 

log  sin  A    =9.7259842 
log  cos  h     =  9.9112682 

log  cos  B    =9.6372524 


Ex.  2.    Given  c  =  72°  30',  o  =  45°  15' ;  find  h,  A,  B. 

Ans.  6  =  64°  42' 52",  A  =  48°  7' 44".5,  B  =  71°  27' 15". 

V 

203.  Case  III. — Qiven  a  side  a  and  the  adjacent  angle  B ; 
to  find  A,  h,  c. 

By  (10),  (6),  (4)  of  Art.  185,  we  have 

cos  A  =  cos  a  sin  B,  tan  6  =  sin  a  tan  B,  cot  c  =  cot  a  cos  B. 

Check  formula,  cos  A  =  tan  b  cot  c. 

In  thill  case  there  is  evidently  no  ambiguity. 

Ex.  1.    Given  a=31°20'45",  B=55°30'30";  find  A,  b,  c. 

Solution. 


log  cos  a  =9.9314797 
log  sin  B  =  9.9160371 


log  cos  A  =  9.8475168 

.-.  A  =  45°  15' 30".6. 

log  cot  a  =0.2153073 
log  cos  B  =  9.7530361 

log  cote   =9.9683434 

.♦.   c  =47°  5' 11". 


log  sin  a  =9.7161724 
log  tan  B  =  0.1630010 

log  tan  6  =9.8791734 

.-.  b  =37°  7' 50". 

Check. 

log  tan  6  =9.8791734 
log  cote   =9.9683434 

log  cos  A  =  9.8475168 


CASE  IV. 


301 


0617 
9325 

9842 

'  48".l. 


19842 
L2682 

r2524 

B. 

°  27'  15". 

angle  B ; 


t  a  cos  B. 


d  A,  b,  c. 

61724 
130010 

91734 


7'  50". 


^91734 
)83434 

175168 


Ex.  2.    Given  o  =  112°  0'  0",  B  =  152°  23'  1".3;  find  A,  6,  c. 
Ans.  A  =  100°,  b  =  154°  7'  26".5,  c  =  70°  18'  10".2. 

204.  Case  IV.  —  Given  a  side  a  and  the  opposite  angle  A  ; 
to  find  b,  c,  B. 


By  (7),  (3),  (10)  of  Art.  185,  we  have 

.    ,      .  ,  .       .  sin  a 

sin  0  =  tana  cot  A,  sinc  = 


.       -r,  COS  A 

~ ,  sin  B  = < 

sin  A  COS  a 


Check  formula, 


smb  =  sine  sin  B. 


In  this  case  there  is  an  ambiguity,  as  the  parts  are  deter- 
mined by  their  sines,  and  two  values  for  each  are  in  general 
admissible.  But  for  each  value  of  6  there  will,  in  general, 
be  only  one  value  for  c,  since  c  and  b  are  connected  by  the 
relation  cos  c  =  cos  a  cos  6  (Art.  185);  and  at  the  same  time 
only  one  admissible  value  for  B,  since  cos  c  =  cot  A  cot  B. 
Hence  there  will  be,  m  general,  only  two  triangles  having 
the  given  parts,  except  when  the  side  a  is  a  quadrant  and 
the  angle  A  is  also  90°,  in  which  case  the  solution  becomes 
indeterminate. 

It  is  also  easily  seen  from  a  figure  that  the  ambiguity 
must  occur  in  general  (Art.  188). 

When  a  =  A,  the  formulae,  and  also  the  figure,  show  that 
b,  c,  and  B  are  each  90°. 

Ex.  1.   Given  a  =  46°  45',  A  =  59°  12' ;  find  b,  c,  B. 

Solution. 

log  tan  a  =0.0265461 
log  cot  A  =9.7753334 

log  sin  b 


9.8018795 

•.  b  =39°19'23".5, 
or  140°40'36".6. 


log  sin  a  =9.8623526 
log  sin  A  =9.9339729 


log  sine    =9.9283797 
.-.  e    =57°  59' 29", 
or  122°  0'31". 


V    '.?( 


'!''* 


!i 


■  I 


302 


SPHERICAL   TlilGONOMETItr. 


log  cos  A  =9.7093063 
log  cos  a  =9.8358066 

log  sin  B  =  9.8734997 

.-.  B  =48°21'28", 
or  131°  38' 32". 


Check, 

log  sine    =9.9283797 
log  sin  B  =  9.8734997 

log  sin  6   =9.8018794 


Ex.  2.   Given  a=  112°,  A  =  100°;  find  6,  c,  B. 
Ans.  b=15i°  7'26".o,    c  =  70°18'10".2,   B  =  152°23'    1".3, 
or     25°52'33".5,   or   109°41'49".8,    or     27°  36' 58". 7. 

205.  Case  V.  —  Oiven  the  two  sides  a  and  b ;  to  find  A, 
B,  c. 

By  (7),  (6),  (1)  of  Art.  185,  we  have 

cot  A  =  cot  a  sin  b,  cot  B  =  cot  b  sin  a,  cos  c  =  cos  a  cos  6. 

Check  formula,  cos  c  =  cot  A  cot  B. 

In  this  case  there  is  no  ambiguity. 

Ex.  1.    Given  a  =  54°  16',  6  =  33°12';  find  A,  B,  c. 

Ans.  A  =  68" 29' 53",  B=  38° 52' 26",  c  =  60°44'46". 
Ex.  2.    Given  a  =  56°  34',  6  =  27°  18';  find  A,  B,  c. 

Ans.  A  =  73°  9'  1 3",  B  =  31°  44'  9",  c  =  60°  41'  9". 

206.  Case  VI.  —  Oiven  the  two  angles  A  and  B ;  to  find  a, 
b,  and  c. 

By  (10),  (9),  (8) 

cos  A    ,   cos  B        i.  A   i.  -D 
cos  a  = ,  cos  0  = ,  cos  c  =  cot  A  cot  B. 

sin  B  sin  A 

Check  formula,        cos  c  =  cos  a  cos  b. 
There  is  no  ambiguity  in  this  case. 

Ex.  1.    Given  A  =  74°  15',  B  =  32°  10' ;  find  a,  b,  c. 

Ans.  a  =  59°  20'  44",  b  =  28°  24'  54",  c  =  63°  21'  24".5. 

Ex.2.    Given  A  =  91°  11',  B  =  111°  11';  find  a,  6,  c. 

Ans.  a  =  91°  16' 8",  6  =  111°  11' 16",  c  =  89°  32' 29". 


EXAMPLES. 


303 


;.S797 
U097 

L8794 


23'    1".3, 

'36' 58".  7. 

0  find  A, 


s  a  cos  b. 


\,c. 


60°44'46". 

?,  c. 
00°  41' 9". 

;  to  find  a, 
.tB. 


b,  c. 

3°  21'  24".r>. 

r,  b,  c. 

89°  32'  29". 


207.  Qnadrantal    and    Isosceles    Triangles.  —  Since    the 

polar  triangle  of  a  quadrantal  triangle  is  a  right  triangle 
(Art.  184),  we  have  only  to  solve  the  polar  triangle  by  the 
I'orniulse  of  Art.  185,  and  take  the  supplements  of  the  parts 
thus  found  for  the  required  parts  of  the  given  triangle ;  or 
we  can  solve  the  quadrantal  triangle  immediately  by  the 
formulae  of  Art.  189.* 

A  biquadrantal  triangle  is  indeterminate  unless  either 
the  base  or  the  vertical  angle  be  given. 

An  isosceles  triangle  is  easily  solved  by  dividing  it  into 
two  equal  right  triangles  by  drawing  an  arc  from  the  vertex 
to  the  middle  of  the  base. 

The  solution  of  triangles  in  which  a  +  &  =  tt,  or  A  +  B  =  ir, 
can  be  made  to  depend  on  the  solution  of  right  triangles. 
Thus  (see  the  second  figure  of  Art.  191)  the  triangle  K'AC 
has  the  two  equal  sides,  «'  and  b,  given,  or  the  two  equal 
angles,  A  and  B',  given,  according  as  a  +  6  =  tt  or  A  -f  B  =  tt 
in  the  triangle  ABC. 

EXAMPLES.  •;     . 

Solve  the  following  right  triangles  :  ' 


1. 


3. 


5. 


Given  c=32°34', 
find  a=22°15'43". 

Given  c= 69°  25' 11", 
find  a=50°  0'  0", 

Given  c=55°  9' 32", 
find   6 =51°  53', 

Given  c=127°12', 
find   &=39°  6' 25", 

Given  a.=  118°  54', 
find  A =95°  55'  2", 


a  =  44°  44'; 
&=24°24'19", 

A=54°54'42"; 
6 =56°  50' 49", 

«=22°15'  7"; 
A=27°28'37'.5, 

a=141°ll'; 

A =128°  5' 54", 

B  =  12°19'; 
6= 10°  49' 17", 


B=50°8'21". 


B  =  63°25'4". 


B=73°27'll".l. 


B  =  52°  21 '49". 


c= 118°  20' 20". 


*  Quadrantal  triangles  are  generally  avoided  in  practice,  but  when  unavoidable, 
they  are  readily  aolved  by  either  of  these  method*. 


I'' 


304 


0. 


SPHERIC  A  L   TRW  ON  OMETR  Y. 


7. 


8. 


9. 


10. 


11. 


12. 


Given  a =29° 46'  8", 
find  A= 54°   1'16", 

Given  a= 77° 21' 50", 
find  6=28°14'31".l, 
or     //=151°45'28".9, 

Given  a =68°, 
find   6=25°52'33".5, 
or     6'=154°7'26".5, 

Given  a  =  144°  27' 3", 
find  A  =  126°  40' 24", 

Given  a =36°  27', 
find  A  =  46°59'43".3, 

Given  A  =  63°  15' 12", 
find  a. =49°  59' 56", 

Given  A  =  67°  54' 47", 


B  =  137°24'2r'; 
6=155°27'64". 

A =83°  56' 40"; 
c= 78°  53' 20", 
c'=101°6'40", 

A  =  80°; 
c=70°18'10".2, 
c'=109°41'49".8, 

&=32°8'56"; 
B=47°13'43", 

6=43°32'31"; 
B=57°59'19".2, 

B  =  135°  33' 39"; 
6=143°5'12", 

B  =  99°  57' 35"; 
&=100°45', 


c=142°9'13". 

B  =28°49'57".4, 
B'=151°10'2".6. 

B  =27°36'58".7, 
B'=152°23'l".3. 

c= 133°  32' 26". 

c=54°20'. 

c=120°55'34". 

c=94°5'. 


find  a= 67°  33' 27", 

13.  Solve  the  quadrantal  triangle  in  which 

c  =  90°,  A  =  42°l',  B  =  121°20'. 
Ans.  C  =  67°16'22",  6  =  112°10'20",  a  =  46°31'30". 

14.  Solve  the  quadrantal  triangle  in  which 

a  =  174°  12'  49".l,  h  =  94°  8'  20",  c  =  90°. 
Arts.  A  =  175°57'10",  B  =  135°42'55",  C  =  135°34'8". 


SOLUTION  OF  OBLIQUE  SPHERICAL  TRIANGLES. 

208.  The  Solution  of  Oblique  bpherical  Triangles  presents 
Six  Cases ;  as  follows :  - 

I.    Given  two  sides  and  the  included  angle,  a,  b,  C. 
II.    Gfiven  two  angles  and  the  included  side,  A,  B,  c. 
III.    Given  two  sides  and  an  angle  opposite  one  of  them, 
a,  6,  A. 


CASE  I. 


306 


2°  9' 13". 

i°49'57".4, 
51°10'2".0. 


7°36'58".7, 
52''23'1".3. 

L33°32'26". 

54°  20'. 
120°  55' 34". 

p94°5'. 


=  46°  31*  3b". 


=  135°  34' 8". 
NGLES. 

ngles  presents 

b,  C. 
B,  c. 
;  one  of  them, 


IV.  Given  two  angles  and  a  side  opposite  one  of  them, 
A,  B,  a. 

V.    Given  the  three  sides,  a,  b,  c. 
VI.    Given  the  three  angles,  A,  B,  C. 

These  six  cases  are  immediately  resolved  into  three  pairs 
of  cases  by  the  aid  of  the  polar  triangle  (Art.  184). 

For  when  two  sides  and  the  included  angle  are  given, 
and  the  remaining  parts  are  required,  the  application  of 
the  data  to  the  polar  triangle  transforms  the  problem  into 
the  supplemental  problem :  given  two  angles  and  the 
included  side,  to  find  the  remaining  parts. 

Similarly,  cases  III.  and  IV.  are  supplemental,  also  V. 
and  VI. 

The  parts  are  all  positive  and  less  than  180°  (Art.  182). 
The  attention  of  the  student  is  called  to  Art.  199. 

209.  Case  I.  —  Oive  i  two  sides,  a,  b,  and  the  included 
angle  C  ;  to  find  A,  B,  c. 

By  Napier's  Analogies,  (7)  and  (8)  of  Art.  197, 


tan  ^  (A  +  B)  = 


-  ^^"  ^  («  -  ft)  cot 


cos  ^  (a  +  6) 


C 

_— • 

2 


,       i/A       T>\      sini^(a  — &)      ,C 
tan  i  (A  -  B)  =   .    \)       ,(  cot-, 
sin  ^  (a  +  6)        2 

These  determine  ^(A  +  B)  and  ^(A  — B),  and  there- 
fore A  and  B ;  then  c  can  be  found  by  Art.  190,  or  by  one 
of  Gauss's  equations  (Art.  198).  Since  c  is  found  from  its 
sine  in  Art.  190,  it  may  be  uncertain  which  of  two  values  is 
to  be  given  to  it :  if  we  determine  c  from  one  of  Gauss's 
equations,  it  is  free  from  ambiguity.  We  may  therefore 
find  c  from  (3)  of  Art.  198.     Thus 

cos  ^  (A  -)-  B)  cos-  =  cos  i  (a  +  &)  sin-- 
2  ^ 


lii 


I  It- 1 


306 


SPHERICAL   TRIGONOMETRY. 


1 

I 

M 

11 

Check,  tan  ^  (a  +  &)  cos  ^  (A  +  B)  =  cos  ^  (A  —  B)  tan  -• 
There  is  no  ambiguity  in  this  case. 

Ex.1.  Given  a  =  43°  18',   &  =  19°24',    C  =  74°22';    find 
A,  B,  c. 

Solution. 

i  {a  4-  h)  =  31°  21',  ^  (ct  -  6)  =  11°  57',  \  C  •-  37°  11'. 


logcosH«-&)=  9-9904848 
log£eci(a  +  &)=  0.0685395 

log  cot  5:^10.1199969 

^       2 

log  tan^(A+B)  =10.1790212 

.-.  ^(A4-B)  =  56°29'17" 
i(A-B)  =  27°41'  0".5 

.-.  A  =  84°10',17".5 

B  =  28°  48' 16".5. 
c  =  41°35'48".5. 


log  sin  |(a  -  6)  =  9.3160921 
iGgcosec^(a+6)=  0.2837757 

logcot- =10.1199969 
^   2 

log  tan  \  ( A  -B)  =  9.7198647 

.-.  i(A-B)=27°41'0".5. 

log  cos  \{a  +  h)=  9,9314605 
log  sec  i(A  4-  B)  =  0.2579737 

log  sin -=  9.7813010 

^        2 


log  cos -=  9.9707352 
^        2 

^  =  20°  47'  54".25. 

2 


Othei'wise  tkus:  Let  fall  the 
perpendicular  BD,  dividing  the 
triangle  ABC  into  two  right  tri- 
angles, BDA,  BDC.  Denote  AD 
by  m,  the  angle  ABD  by  <^,  and  A-^^ 
BDhyp.   Then  by  Napier's  Rules,  b 

we  have 

cos  C  =  tan  (6  —  m)  cot  a ; 

sin  (&  —  m)  =  cot  C  tanp ;  sin  m  =  cot  A  tanp. 

.'.  tan(6  —  m)  =  tanacosC (1) 

and        tan  A  sin  m  =  tan  C  sin  (&  —  m) (2) 


CASE  11. 


307 


ban 


';    find 


11'. 

3160921 

2837757 

1109969 

,7198647 
'0".5. 

,9314605 
1.2579737 

1.7813010 
).9707352 


np. 


.     .     (1) 


From  (1)  m  is  determined,  and  from  (2)  A  is  determined. 
In  a  similar  manner  B  may  be  found. 

Also,  from  the  same  triangles,  we  have  by  Napier's  Rules 

cosa"=  cos  (6  —  m)  cosp;  cos  c  =  cos  m  cosp. 

.•.  cosccos  (6  —  m)  =  cosmcosa, 

from  which  c  is  found. 

Note.  —  This  method  has  the  advantage  that,  in  using  it,  nothing  need  be  remem- 
bered except  Napier's  Rules. 

If  only  the  side  c  is  wanted,  it  may  be  found  from  (4)  of  Art.  191,  without  pre- 
vi3usly  determining  A  and  B.  This  formula  may  be  adapted  to  logarithms  by  the 
nae  of  a  subsidiary  angle  (Art.  90). 

Ex.2,  Given  6=120°30'30",  c=70°20'20",  A=50°10'10"; 
find  B,  C,  a. 

A71S.  B  =:  135°  5'  28".8,  C  =  50°  30'  8".4,  a  =  69°  34'  56". 

210.  Case  11.  —  Given  two  angles,  A,  B,  and  the  included 
side  c ;  to  find  a,  b,  C. 

By  Napier's  Analogies  (9)  and  (10)  of  Art.  197, 

cos^(A  +  B)        2 

:    tanH«-&)='^"^^^~?^-ian£,  ' 

^^         ^      sin^(A  +  B)        2 

from  which  a  and  b  are  found. 
The  remaining  part  C  may  be  found  by  (2)  of  Art.  198. 

sin  ^  (a  —  &)  cos  -  =  sin  ^  (A  —  B)  sin -• 

n 

Check,      3s  ^  (a  —  b)  cot  —  =  cos  ^  (a  -f  b)  tan  ^  (A  +  B). 

There  is  no  ambiguity  in  this  case. 


ti 


m,f 


■•■•V- 


■:,;■■  t' 


308 


SPHEltlCAL   TRIGONOMETRY. 


Ex.  1.  Given  A=68°40',  B  =  56°20',  c=84°30';  find  a,  b,  C. 

Solution. 
i  (A  +  B)  =  62°  30',  i  (A  -  B)  =  6°  10',  ^  =  42°  15'. 


logcosi(A-B)=  9.9974797 
logseci(A+B)=  0.3355944 

log  tan  ?=  9.9582465 
^       2     . . 


log  tan  |(a  +  6)  =  10.2913206 

.-.  i(a.  +  &)  =  62°55'   9" 

^(a-b)=   6°  16' 39" 

a  =  69°  11'  48" 

b  =  56°  38'  30". 
C=::97°19'    3".5. 


log  sin  ^(A  -  B)   =9.0310890 
logcoseci(A  +  B)  =0.0520711 

logtan- =9.9582465 
^        2     

log  tan  |(a  -  &)  =9.0414066 

.-.  ^(a- 6)  =6°  16' 39". 

log  sin  |(  A  - B )  =  9.0310890 
log  cosec  ^(a  -  b)  =0.9612050 

log  sin  £=9.8276063 

^       2     

log  cos  ^=9.8199003 
.  §  =  48°39'31f". 


Otherwise  thus :  Let  fall  the  perpendicular  BD  (see  last 
figure).  Denote,  as  before,  AD  by  w,  the  angle  ABD  by  </>, 
and  BD  by  p.     Then  by  Napier's  Rules,  we  have 

cos  c  =  cot  ^  cot  A ; 

cos  (^  =  cote  tan ^;  cos(B  —  <^)  =  cotatanp. 

.*.  cot  <^  =  tan  A  cos  c (1) 

tan  a  cos  (B —  </>)=  cos  (^  tan  T (2) 

From  (1)  <^  is  determined,  and  from  (2)  a  is  found. 
Similarly  6  may  be  found. 

Also,  from  the  same  triangles,  we  have 

cos  C  sin  (f>  =  cos  A  sin(B  —  </>), 
from  which  C  is  found. 


CASE  III. 


309 


a,  b,  C. 

15'. 

)310890 
0520711 

9582465 

0414066 
'  16'  39". 

,0310890 
.9612050 

.8276063 
1.8199003 
If". 

(see  last 
BD  by  </>, 


IP- 

.     (1) 

•     (2) 

IS  found. 


Ex.  2.    Given 
A  =  135°  5'  28".G,  C  =  50°  30'  8".6,  b  =  69°  34'  56".2 ; 
find  a,  c,  B. 

Ans.  a  =  120°  30' 30",  c  =  70°  20' 20",  B  =  50°  10' 10". 

21L   Case  III.  —  Given  two  sides,  a,  b,  and  the  angle  A 
opposite  one  of  them ;  to  find  B,  C,  c. 

The  angle  B  is  found  from  the  formula, 

sinB  =  ^HL^sinA (Art.  190)  (1) 

sin  a 

Then  C  and  c  are  found  from  Napier's  Analogies,        ' 

tang  =  ^^"i(^-^)cotHA-B)  .     .     . 
2      sini(a  +  6)        ^^  ^ 

•       :       '        c       sini( A  +  B)  ,      ,  /        ,v 

tan- = -^ — "^     ^ — ^tanH«  — ^)-     •     • 
2      sini-(A-B)        ^^  ' 


(2) 
(3) 


Check, 


sin  A 
sin  a 


sinB 
sin  b 


sinC 


sin  c 


Since  B  is  found  from  its  sine  in  (1),  it  will  have  two 
values,  if  sin  A  sin  6  <  sin  «,  and  the  triangle,  in  general, 
will  admit  of  two  solutions. 

When  sin  A  sin  b  >  sin  a,  there  will  be  no  solution,  for 
then  sin  B  >  1. 

In  order  that  either  of  these  values  found  for  B  may  be 
admissible,  it  is  necessary  and  sufficient  that,  when  sub- 
stituted  in    (2)    and    (3),   they   give   positive  values   for 

C  c 

tan—  and  tan-,  or  which  is  the  same  thing,  that  A  —  B  and 

Z  Li 

a  —  b  have  the  su,me  sign.     Hence  we  have  the  following 

Rule.  —  If  both  values  of  B  obtained  from  (1)  be  such  as 
that  A  —  B  and  a  —  b  have  like  signs,  there  are  tivo  complete 
solutions.  If  only  one  of  the  values  of  B  satisfies  this  condi- 
tion, there  is  only  one  triangle' that  satisfies  the  problem,  since 


]'i-m 


810 


SPHERICAL   TRIGONOMETRY. 


'^i'^- 


in  this  case  C,  or  c  >  180°.  If  neither  of  the  values  of  B 
makes  A  —  B  and  a  —  b  of  the  same  signs,  the  problem  is 
impossible. 

This  case  is  known  as  the  ambiguous  case,  and  is  like  the 
analogous  ambiguity  in  Plane  Trigonometry  (Art.  116), 
though  it  is  somewhat  more  complex.  For  a  complete 
discussion  of  the  Ambiguous  Case,  the  student  is  referred 
to  Todhunter's  Spherical  Trigonometry,  pp.  53-58 ;  McCol- 
lend  and  Preston's  Spherical  Trigonometry,  pp.  137-143; 
Serret's  Trigonometry,  pp.  191-195,  etc. 

Ex.  1.  Given  a=42°45',  &=47°15',  A=56°30';  findB,  C,c. 


Solution. 


log  sin  6  =  9.8658868 

colog  sin  a  =  0.1682577 

log  sin  A  =  9.9211066 

log  sin  B  =9.9552511 
.-.  B  =  64°26'4", 
B'  =  115°  33'  56". 


^{a-hb)  =  45°  0'  0". 
^(a-b)  =-  2°  15'  0". 
■^(A  +  B)  =  60°  28'  2". 
|(A  _  B)  =  -  3°  58'  2". 
|(A  +  B')=  86°  1'58". 
|(A-B')  =  -29°31'58". 


Since  both  values  of  B  are  such  that  A  —  B,  A  —  B',  and 
a  —  b,  are  all  negative,  there  are  two  solutions,  by  the  above 
Rule. 


(1)  TT/iew  B  =  64°  26' 4". 

logsini(a-6)   =8.5939483- 
cologsin^(a+&)  =0.1505150 
log  cot  ^( A  -  B)  =  1.158941.'.- 

log  tan  ^=9.9034046 
®        2 


C 


=  38°  40'  48". 


.-.  C  =  77°  21' 36". 


log  sin  i(  A  +  B)  =9.9395560 
cologsin^(A-B)=1.1599832- 
log  tan  i{a-b)    =  S.5942832 - 

log  tan -=9.6938224 
^        2 

.-.  -=26°  17' 40". 
2 

.-.  c  =53°  35' 20". 


CASE  III. 


311 


es  of  B 
Mem  is 

like  the 
t.  IIC), 
jomplete 
referred 

;   McCol- 

137-143; 


idB,C,c. 

r  0'  0". 
2°  15'  0". 
3°  28'  2". 
3°  58'  2". 
G°  1'58". 
9°  31'  58". 

—  B',  and 
the  above 


9.93955G0 
1.1599832- 

8.5942832- 

=9.6938224 

=2G°17'40". 
=  53°  35' 20". 


(2)  When  B'  =  115°  33'  56". 

log  sin  ^(a  -  b)   =8.5939483- 
cologsin|(rt+ft)  =0.1505150 
log  cot  ^(A-B')  =0.2467784- 

log  tan  ^-=8.9912417 
^        2 


C' 

2 


5°35'50i". 


•.  C'=11°11'40^". 


logsin|(A  +  B')   =9.9989581 
colog  sin  |(  A -B')  =0.3072223- 
log  tan |(a  -  ^;)  =  8.5942832- 

log  tan  ^^'=8.9004636 

^        2 

....  ^' =  4°  32' 47i". 

2  * 

.-.  c'=9°    5' 34V'. 


Ans.  B  =    64°  26'   4",  C  =77°  21' 36",     c  =53°  35' 20"; 
B' =  115°  35' 56",  C'  =  11°11'40V',  c'=   9°    5' S^". 

Otherwise  thus:    Let  fall  the  ' 

perpendicular  CD;  denote  AD 
by  m,  the  angle  ACD  by  <fi,  and 
CD  by  p.     Then  we  have 

cos  A  =  tan  m  cot  6 ;  ^ 

.'.  tan  m  =  cos  A  tan  b      (1) 

cos  b  =  cot  A  cot  <l>.     .  •.  cot  (f}  =  cos  b  tan  A 
Again,  .     ,^ 

cos  a  =  cos  (c  —  m)  cosp ;  cos  6  =  cos  m  cosjp. 
.•.  cos  (c  —  m)  =  cos  a  cos  7Ji  -i-b    . '  . 
Also,  cos  (C  —  <^)  =  cot  a  tsaip ;  cos  <^  =  cot  b  tajip. 
.'.  cos  (C  —  <^)  =  cot  a  tan  6  cos  </>  .     . 


(2) 


(3) 


Lastly, 


.    T3      sin  6    .     . 
sin  B  =  -^ —  sin  A 

sin  a 


(4) 

(5) 


The  required  parts  are  given  by  (1),  (2),  (3),  (4),  (5). 

Ex.  2.  Given  a  =  73°49'38",  ?;=120°53'35",  A=88°52'42": 
find  B,  C,  c. 

Ans.  B  =  116°  44'  48",  C  =  1 16°  44'  48",  c  =  120°  55'  35". 


:</  i^ 


312 


BPHEJilCAL  TRIGONOMETRY. 


212.  Case  IV.  —  Given  two  angles,  A,  B,  and  the  side  a 
opposite  one  of  them ;  to  find  b,  c,  C. 

This  case  reduces,  by  aid  of  the  polar  triangle,  to  the 
preceding  case,  and  gives  rise  to  the  same  ambiguities. 
Hence  the  same  remarks  made  in  Art.  211  apply  in  this 
case  also,  and  the  direct  solution  may  be  obtained  in  the 
same  way  as  in  Case  III.     Thus, 

The  side  b  is  fo'und  from  the  formula 

sin  B 


sin& 


sin  a (1) 


sin  A 
Then  c  and  C  are  found  from  Napier's  Analogies. 

2      cosi(A-B)        ^^    ^    ^ 

tang  =  ^^^^(^-^-)coti(A  +  B)      .     . 
2      cos  ^  (a +  6)       ^^  ^ 


(2) 
(3) 


Cfheck, 


sin  A     sin  B      sin  C 


sin  a 


sin  6 


sine 


Ex.  1.   Given  A=GG°7'20",  B=52°50'20",  a=59°28'27"; 

find  6,  c,  C. 

Solution. 

By  (1)  we  find  6  =  48°  39' 16",  or  131°  20' 44". 


|(A4-B)=59°28'50". 
^(A-B)=  6°  38' 30". 

log  cos  |(A  4-  B)  =  9.7057190 

cologcosi(A-B)=  0.0029244 

log  tan  i  (a  +  b)  =  10.1397643 


log  tan -=  9.8484077 

^        2 

J  =  35°ll'50f". 


|(a  +  6)  =  64°   3'51i". 
|(a-&)=    r)°24'35.i". 

log  cos  ^  (o  -  b)  =9.9980612 
cologcos^(a  +  &)  =0.2314530 
log  cot  ^(A  +  B)  =9.7704854 

log  tan  ^  =9.9999996 


C 

2 


=  45°. 


The  second  value  of  6  is  inadmissible  (see  Rule  of  Art. 
211),  and  therefore  there  is  only  one  solution. 

Ans.  h  =  48°  39'  16",  c  =  70°  23'  41^",  C  =  90°. 


CASE  V. 


313 


side  a 

to  the 
juities. 
in  this 
in  the 


.     (1) 

.     (2) 
.     (3) 


•28' 27"; 


!'  51^". 
t'35.J". 

).9980G12 
).2314530 
3.7704854 

9.9999996 


3  of  Art. 
,  C  =  90°. 


Ex.  2.  Given  A  =  110°  10',  B  =  133°  18',  a  =  147°  5'  32" ; 
find  b,  c,  C. 

Ans.  b  =  155°  5'  18",  c  =  33°  1'  45",  C  =  70°  20'  50". 

213.  Case  V.  —  Given  the  three  sides,  a,  b,  c,  to  find  the 
angles. 

The  angles  may  be  computed  by  any  of  the  formulae  of 
Art.  195 ;  but  since  an  angle  near  90°  cannot  be  accurately 
determined  by  its  sine,  nor  one  near  0°  by  its  cosine  (Art. 
151),  neither  of  the  first  six  formulae  can  be  used  with  advan- 
tage in  all  cases.  The  formulae  for  the  tangents  however  are 
accurate  in  all  parts  of  the  quadrant,  and  are  therefore  to 
be  preferred  for  the  solution  of  a  triangle  in  which  all  three 
sides  or  all  three  angles  are  given. 

By  (7)  of  Art.  195  we  have 


tan 


_  •  /sin  (s  —  b)  sin  (g  —  c)        . 
\      sins  sin  {s  —  a) 

1  /sin  {s  —  a)  sin  {s  —  b)  sin  {s  —  c) 

sin(s  — a)\  sins 

Since  the  part  under  the  radical  is  a  symmetric  function 
of  the  sides,  it  is  in  the  formulae  for  determining  all  three 
angles  A,  B,  C,  and  when  once  calculated,  it  may  be  utilized 
in  the  calculation  of  each  angle.  Foi;  convenience  in  com- 
putation, denote  this  term  by  tanr.     Then 


tan?' 


=V^ 


sin  (s  — -  g)  sin  {s  —  b)  sin  {s  —  c) . 


sms 


and  (7),  (8),  (9)  of  Art.  195  become 


tanf 
tan| 
tan| 

tanr 

sin  (s  —  a) 
tanr 

sin  (s  —  b) 

tan  r 

amis  —  c) 

(1) 

(2) 
(3) 


314 


SPHERICAL   rniGONOMETRY. 


Check, 


sin  A  _  sin  B  _  sinC^ 
sin  a       sin  b       sin  c 


Ex.1.   Given    a  =  4C°  24',   &  =  G7°14',   c  =  81°12';    fkid 


A,  B,  C. 


Solution. 


a=  46^24' 
6=  67°  14' 
c=   81°  12' 


2  s  =  194°  50' 

s=  97°  25', 

8-a=  51°    r, 

8-6=  30°  11', 

s-c=  16°  13'. 


log  sin  (s  -  a)  =  9.8906049 

log  sin  {s  -  b)  =  9.7013681 

log  sin  {s  -  c)  =  9.4460251 

colog  sin  3  =  0.0036487 

log  tan2r  =  9.0416468 

log  tan  r  =  9.5208234. 


tan  r  =9.5208234 
sin(s-a) =9.8906049 


tan -=9.6302185 
2 

.-.  4=23°    6' 45". 
2 

A  =46°  13' 30". 


tan  r  =9.5208234 

sin  (s-/>)  =  9.7013681 


tan -=9.8194553 

2 

•.  ?  =33°  25' 10". 
2 

B  =66°  50' 20". 


tanr=  9.5208234 
sin(s-c)=  9.4460251 

tan  ^=10.0747983 


C 


:49°54'35 


»  QK" 


C  =99°  49' 10". 


.-<. 


Ex.2.   Given  a  =  100°,  &  =  37°  18',  c  =  62°46';    find   A, 
B,  C. 

Ans.  A=176°15'46".56,  B  =  2°17'55".08,  C=3°22'25".46. 

214.  Case  VI.  —  Given  the  three  angles,  A,  B,  C ;  to  find 
the  sides. 

As  in  Art.  213,  the  formulae  for  the  tangents  are  to  be 
preferred. 

Putting    tan  R  =^^^^  ^^  _  ^^  ,,"  |i:|y^^^> 

we  have,  from  (7),  (8),  (9)  of  Art.  196, 

tan  -  =  tan  R  cos  (S  —  A), 
2t 


CASE  VI. 


315 


tan-  =  tan R  cos  (S  —  B), 

tan  I  =  tan  R  cos  (S  —  C), 

by  which  the  three  sides  may  be  found. 

rn    7                 sin  A      sin  B      sin  C 
VhecK,  = = • 

sin  a      sm  h      sin  c 
Ex.  1.    Given   A  =  68°  30',  B  =  74°  20',  C  =  83°  10';   find 

Solution. 


a,  b,  c 


A=    68°  30' 

B=    74°  20' 

C=    83°  10' 

28  =  226°   0' 

log  (- cos  S)  =  9.5918780 

log  cos  (S  -  A)  =  9.8532421 

log  cos  (S  -  B)  =  9.8925365 

log  cos  (S  -  C)  =  9.9382576 

log  tan2R  =  9.9078418 

log  tan  R  =  9.9539209.* 


S  =  113°   0' 
S  _  A  =   44°  30'  . 

S-B=   38°  40' 

S  -  C  =    29°  50' 

lt)g  tan -  =  9.8071630 

^        2 

log  tan  -  =  9.8464574 
^        2 

log  tan -  =  9.8921785 

^        2      


a  =  65°  21'  22^" 
6  =  70°  9'  9^', 
c  =  75°55'    9". 


Check, 


sin  a  _  sin  b  _  sine  ^ 
sin  A     sin  B     sin  C 


Ex.2.  Given  A=59°55'10",  B=85°36'50°,  C=59°55'10"; 
find  a,  b,  c. 

Ans.  a  =  129°  11'  40",  b  =  63°  15'  12",  c  =  129°  11'  40". 

*  Tlio  neceBsary  additions  may  be  conveniently  performed  by  writing  log  tan  R  on 
a  slip  of  paper,  and  holding  it  Buccessively  over  log  cos  (8  -A),  log  cos  (S  —  B),  and 

l0gC08(S-C). 


t»l  .' 


••♦ 


iW' 


316 


SPHERICAL   TRIGONOMETRY. 


EXAMPLES. 

Solve  the  following  right  triangles : 

Given  c=  84° 20', 
find   a=  35°  13'  4", 

Given  c=  G7°54', 
find   a=  39°35'i)l", 

3.  Given  c=  22°  18' 30", 
find   a=  1G°17'41", 

4.  Given  c=145°, 
find  a=  13°  12' 12", 

6.   Given  c=  98°   6' 43", 
find   a =137°  6', 

6.  Given  c=  4G°40'i2", 
find   a=  26°27'23".8, 

7.  Given  c=  7C°42', 
find    b=  70°  10' 13", 

8.  Given  c=  91°  18', 
find    b=  94°18'53".8, 

Given  c=  86° 51', 
find  a=  86° 41' 14", 


10. 


11. 


12. 


13 


14 


Given  c=  23° 49' 51", 

find   b=  19°  17', 

Given  c=  97°  13'  4", 

find   6=  79°13'38".2, 

Given  c=  37° 40' 20", 

find    6=  0°2G'37".2, 


A=  35°  25'; 

6=  83°  3' 29", 

B=  85°  59'  1". 

\  15 

A=  43°  28'- 

b=  60°46'25f', 

B=  70°  22' 21". 

A=  47°  39' 36"; 

16. 

b=  15°  26' 53", 

B=  44°33'53".4, 

A=  23°  28'; 

17. 

6= 147°  17' 15", 

B  =  109°  34' 33". 

A=138°27'18"; 

18. 

6=  77°  51', 

B=  80°  55' 27". 

A=  37°  46'  9"; 

10. 

b=  39°57'41".4, 

B=  62°  0'  4". 

a=  47°  18'; 

A=  49°  2'24".5, 

B=  75°9'24".7o. 

20. 

a=  72°  27'; 

A=  72''-'29'48", 

B=  94°  6'53".3, 

b=  18°   1'50"; 

21. 

A=  88°  58' 25", 

B=  18°  3' 32". 

a=  14°  16' 35"; 
A=  37°36'49".3, 

B=  54°49'23".3. 

22. 

a=  132°  14' 12"; 

A =131°  43' 50", 

B=  81°58'53".;5. 

23. 

«:=  37°  40' 12"; 

A=  89°  25' 37", 

B=     0°43'33". 

tp 


EXAMPLES. 

317 

13. 

Given  a=  82°  6', 

B=  43°  28'; 

find  A=  84°34'28", 

6=  43°  11' 38", 

c=  84°  14' 57". 

14. 

Given  a=  42°30'30", 

B=  53°  10' 30"; 

find  A=  53°  50' 12", 

b=  42°   3' 47", 

c=  56° 49'  8". 

85°  59'  1".     1 

15. 

Given  a=  20°20'20", 

B=  38°  10' 10"; 

find  A  =  54°35'16".7, 

6=  15°16'50".4, 

c=  25°14'38".2. 

70°  22' 21". 

16. 

Given  a=  92° 47' 32"; 

B=  50°  2'  1"; 

. 

find  A=  92°  8' 23", 

6=  50°, 

c=  91°47'40". 

44°33'53".4. 

17. 

Given  b=  54°  30', 

A=  35°  30'; 

\.    ■ 

109°  34' 33". 

find  B=  70°17'35", 

a=  30°  8'39".2, 

c=  59°51'20". 

18. 

Given  6=155°46'42".7, 

A=  80°  10' 30"; 

:  80°  65' 27". 

find  B=153°58'24".5, 

a=  67°   6'52".6, 

c=110°46'20". 

19. 

Given  a=  35°  44', 

A=  37°  28'; 

b  62°  0'  4". 

find    b=  69° 50' 24", 
or      6'=110°   9'36", 

c=  73°45'15", 
c'=rl06°14'45". 

B=  77°  54', 
B=102°  6'. 

:  75°9'24".7o. 

20. 

Given  rt=129°33', 
find    b=  18°  54' 38", 

A  =  104°  59'; 
c=127°  2' 27", 

B=  23°  57' 19", 

=  94°  6'53".3. 

or      &'=161°  5'22", 

c'=  52°57'33", 

B'=156°  2'41". 

21. 

Given  a=  21°  39', 

A=  42°  10' 10"; 

=  18°  3' 32". 

find    b=  25°  59' 27  ".8, 
or      6'= 154°  0'32".2, 

c=  33°20'13".4, 
c'=146°39'46".6. 

B=  62° 23'  2".8, 
B'=127°36'57".2. 

=  54°  49' 23".;]. 

22. 

Given  a=  42°18'45", 
find    b=  60° 36' 10", 

A=  46°  15' 25"; 
c=  68°  42' 59", 

B=  69°  13' 47", 

or      6' =119°  23' 50", 

c'=lll°17'  1", 

B'=110°46'13". 

=  81°58'53".;J 

23. 

Given  &=160°, 

B  =  150°; 

1 

find   a=  39°  4'50".7, 

c=136°50'23".3. 

A=  67°  9'42".7, 

=     0°43'33".  1 

or     a'=140°55'  9".3, 

c'=  43°  9' 36". 7, 

A'=112°50'17".3. 

s 


R 


1 


318 


24. 


25. 


26. 


27. 


28. 


^o» 


30. 


31. 


32. 


33. 


34. 


35. 


36. 


SPUE  RICA  L   TRIG  ONOMETR  F. 


Given  a= 

25°  18' 45", 

A=  15°  58' 15". 

Ana.  Impossible  ;  why? 

Given  a= 

32°  9' 17", 

b=  32°  41'; 

find  A  = 

49°  20' 17", 

B=  50°  19' 16", 

c=  44°33'17". 

Given  a= 

55°  18', 

b=  39°  27'; 

find  A= 

66°  15'  6", 

B=  45°   1'31", 

c=  63°55'21". 

Given  a= 

56°  20', 

b=  78°40'; 

find  A= 

56° 51'  7", 

B=  80°  31 '48", 

c=  83°44'44i". 

Given  a= 

8'3°40', 

6=  32°  40'; 

find  A  = 

88°11'57".8, 

B=  32°42'37".8, 

c=  87°11'39".S. 

Given  a— 

37°  48' 12", 

b=  59°  44' 16"; 

find  A= 

41°  55' 45", 

B=  70°  19' 15", 

c=  66° 32'  6". 

Given  a= 

116°, 

b=   16°; 

find  A  = 

97°39'24".4, 

B=  17°41'39".9, 

c=114°55'20".4. 

Given  A  = 

52°  26', 

B=  49°  15'; 

find  a= 

36°24'34".5, 

6=  34°  33' 40", 

c=  48°  29' 20". 

Given  A  = 

64°  15', 

B=  48°  24'; 

find   a= 

54°  28' 53", 

6=  42°  30' 47", 

c=  64°  38' 38". 

Given  A  = 

54°  1'15", 

B  =  137°24'21"; 

find  a= 

29°  46'  8", 

6=155°27'55", 

c=142°  9' 12". 

Given  A= 

46°  59' 42", 

B=  57°  59' 17"; 

find   a= 

36°  27', 

6=  43°  32' 37", 

c=  54° 20'  3". 

Given A = 

55°  32' 45", 

B=101°47'56"; 

find   a= 

54°41'35", 

6= 104°  21' 28", 

c=  98°  14' 24". 

Given A = 

60°27'24".3, 

B=  57°16'20".2; 

find  a= 

54°32'32".l, 

b=  51°43'36".l, 

c=  68°  56' 28".  9. 

4°  33' 17". 
>3°55'21". 

33°44'44i". 
87m'39".S. 
66°  32'  6". 
14°55'20".4. 

48°  29' 20". 

64°  38' 38". 

42°  9' 12". 

54°  20'  3". 

«)8°14'24". 
68°56'28".9. 


37. 


38. 


39. 


40. 


41. 


42. 


43. 


44. 


45. 


46. 


47. 


48. 


EXAMPLES. 

ijiy 

Solve  the  following  quaili 

•antal  triangles : 

Given  B=  74°  45', 

a=  18°  12', 

c=   90°; 

find   6^  86°17'15".6, 

A=  17°  34'  2", 

C  =  104°  31' 13". 

Given  A  =  l  10°  47'50", 

B=135°35'34''.5, 

c=  90°; 

find  a =104°  53'  0".8, 

6=133°39'47".7, 

C  =  104°41'37".2. 

Solve  the  following  oblique  triangles: 

Given  a=  73°  58', 

b=  38°  45', 

C=  46°  33' 39"; 

find  A=116°  8' 28", 

B=  35°  46' 39", 

c=  51°   I'll". 

Given  a=  96°  24' 30", 

6=  68°  27' 26", 

C=  84°  46' 40"; 

find  A=  97°53'0i", 

B=  67°59'39i", 

c=  87°  31' 37". 

Given  a=  76° 24' 40", 

b=  58°  18' 36", 

C  =  116°  30' 28"; 

find  A=  63°48'35V', 

B=  51°46'12i", 

c= 104°  13' 27". 

Given  a=  86°18'40", 

b=  45°  36' 20", 

C  =  120°  46' 30"; 

find  A=  64°48'53|", 

B=  40°23'15|", 

c=108°39'llV'. 

Given  a=  88°  24', 

b=  56°  48', 

C  =  128°16'; 

find  A=  65°  13'  3^', 

B=  49°  27' 51", 

c= 120°  10' 52". 

Given  a=  68°20'25", 

6=  52°  18' 15", 

C=117°12'20"; 

find  A=  56°  16' 15", 

B=  45°  4'41", 

c=  96°20'44". 

Given  a=  88°12'20", 

6=124°   7'17", 

C=  50°  2'  1"; 

find  A=  63°  15' 12", 

B= 132°  17' 59", 

c=  59°  4' 25". 

Given  a=  32°  23' 57", 

6=  32°  23' 57", 

C=  66°49'17"; 

find  A=  60°  53'  2", 

B=  60°  53'  2", 

c=  34°  19' 11". 

Given  b=  99° 40' 48", 

c=100°49'30". 

A=  65°  33' 10"; 

find  B=  95°  38'  4", 

C=  97°26'29".l, 

a=  64°23'15".l. 

Given  A  =  31°  34' 26", 

B=  30°  28' 12", 

c=  70°  2'  3"; 

find  «=  40°  1'  5\", 

b=  38° 31'  3^", 

C=130°  3' 50".                       ■ 

320 

49. 

50. 
51. 

52. 

* 

53. 
54. 
65. 
56. 

57. 


SPHERICAL    TRIGONOMETRY. 


58. 


59. 


60. 


Given  A  = 
find   a= 

Given  A  = 
find   a= 

Given  A  = 
find   a= 

Given  Ax 
find   «= 

Given  A  = 
find   a- 

Given  A: 
find   a- 

GivenA: 
find   a- 

GivenA: 
find   a: 

Given  «= 
find  B  = 
or     B': 

Given  a- 
find  B: 
or     B': 

Given  a-. 
find  B- 
or     B': 

Given  a-. 
find  B: 
or     B': 


:130°5'22".4, 
:  84°  14' 29", 

:  96°  46' 30", 
=  102°  21 '42", 

:  84°  30' 20", 
:  94°  34' 52V', 

=  107° 47'  7", 
=  70°  20' 50", 

=  128°  41' 49", 
=  125°  44' 44", 

=  129°58'30", 
=  85°  59', 

=  95°  38'  4", 
=  99°40'48", 

=  70°, 

=   57^  56' 53", 

=  62°  15' 24", 
=  62°24'24".8, 
=  117°35'35".2, 

=  52°  45' 20", 
=  59°24'15|", 
=  120°  35' 44 1", 

=  48°  45' 40", 
=  55°  39' 57", 
=  124°  20'  3", 

=  46°  20' 45",  '  6= 
=  54°  6' 19",  \  C  = 
=  125°53'41",      C'= 


B  = 

h-. 

B= 

6= 

B  = 

B= 

h-. 

B  = 

h-. 

B= 

6= 

C: 

C: 

B: 

6: 

h: 

C: 

C: 

6: 

C: 

C: 

h: 
C: 


!  nf 


C'= 


32°26'6".41, 
44°  13' 45", 

84°  30' 20", 

78°  17'  2", 

76°  20' 40", 
76°  40' 48  V', 

38°  58' 27", 
38°  27' 59", 

107°  33' 20", 
82°  47' 35", 

34°  29' 30", 
47°  29' 20", 

97°  26' 29", 
100°  49' 30", 

131°  18', 
137°  20' 33", 

103°  18' 47", 
155°43'11".3, 
59°  6'10".6, 

71°12'40", 

115°  39' 55V', 
26°59'55".2, 

67°  12' 20", 

116°  34' 18", 

24°  32' 15", 

65°  18' 15", 
116°  55' 26", 
24°12'53".3, 


c= 
C  = 

c= 
C  = 

c= 
C  = 

c- 
C  = 

c= 
C  = 

C-- 
C  = 

h: 
B: 

C- 

C  = 
A  = 

c'= 

A= 


c'= 

A  = 

c= 
c'= 

A  = 

c= 
c'= 


:     51°     6 

:  36° 45 

:  126° 46 
=125° 28 

130° 46 
:  130° 51 

:   51°  41 

:  52° 29 

:  124° 12 
=127° 22 

:   50°   6 
:  36°   6 

:  64° 23 
:  65° 33 

=116°; 
:  94° 48 

53°  42 

153°  9 

70°  25 

46°  22 
97°  33 
29°  57 

42°  20 
93°  8 
27°  37 

40°  10 
90°  31 
27°  23 


ll"/); 
26". 


13V'. 


33V'. 

14"; 

45". 

31"; 

7". 

20"; 
50". 

15"; 
10". 


12". 
38"; 

26". 

10"; 

18".8, 

io".r,. 

30"; 
9".(i, 
20". 

30" ; 
46", 
14". 


19. 


EXAMPLES. 


321 


,1°   6'11".0;    I  61. 
16°  45' 26". 


>6°46';^ 
25°28'13|". 

^0°46'; 
30°  51' 33  V- 


24°  12' 31"; 

.27°  22'  7". 

50°   6' 20"; 
36°   6' 50". 


46°  22' 10"; 
:  97°33'18".8, 
L'0°57'10".r). 


62. 


51°  41' 14";      163. 
52°  29' 45". 


64. 


64°  23' 15";     |g5 
65°  33' 10". 

116° ; 
94°  48' 12". 

53°  42' 38"; 

153°  9' 35V', 
70°  25' 26". 


18. 


42° 20' 30"; 
:  93°  8'  9".<'», 
=  27°  37' 20". 

=  40°  10' 30";    |9 
=  90°  31' 46", 
=  27"  23' 14". 


Given  a=150°57'  5", 
find  B  =  120° 47' 44", 
or    B'=  59°  12' 16", 

Given  a=  50° 45' 20", 
find  B=  57°34'51".4, 
or     B'=122°25'  8". 6, 

Given  a=  40°  5'25".6, 
find  B=  42°  37' 17 ".5, 
or     B'=137°22'42".5, 

Given  a=  99° 40' 48", 
find  B=  65°  33' 10", 
(No  ambiguity  ;  why?) 

Given  A  =  79°  30' 45", 

find    b=  36°  5'34f' , 
(No  ambiguity ;  why?) 

Given  A  =  73°  11' 18", 

find    b=  41°52'34|", 
(Only  one  solution  ;  why?) 

Given  A  =  46°  30' 40", 
find   b=  33°18'47i", 
(Only  one  solution  ;  why?) 

Given  A  =  61°  29' 30", 

find    b=  15°30'30".5, 
(Only  one  solution  ;  why?) 

Given  A  =  36°  20' 20", 
find    6=  55°  25'  2^", 
or     6' =124°  34' 57^", 


b-. 
C  = 
C'= 

6: 
C: 


134°  15' 54", 

:  97°  42' 55", 
:  29°   9'  9", 

:  69°  12' 40", 
:115°57'50".6, 
:  25°44'31".6, 


:  118° 22'   7".3, 
:160°   1'24".4, 


b 
C 
C'=  50°18'55".2,    c'= 


c'= 


A  = 

c= 
c'= 


A  = 


6=  64°  23' 15". 
C=  97°  26' 29", 


B  = 

C-- 

B: 

C: 


B: 


B: 


V.= 


46°  15' 15", 
50°  24' 57", 

61°  18' 12", 
41° 35'  4", 

36"  20' 20", 
60°  32'   6", 

24°  30' 30", 
39°  33' 52", 

46°  30' 40", 
81°  27' 26  V', 


A  =  144°  22' 42"; 

55° 42'  8", 
23°  57' 29". 

44° 22' 10"; 
95n8'16".4, 
28° 45'  5".2. 

29°42'33".8; 

153°38'42".4, 

90°   5'41".0. 

:  95°  38'  4"; 
=  100°  49' 30". 


a=  53°18'20"; 
C=  70°  55' 35". 


a=  46°  45' 30"; 
C=  60°42'46".5. 


a=  42°  15' 20"; 
C  =  110°  3'14".6. 


a=  34°  30"; 
C=  98'48'58".5. 


a=  42°  15' 20"; 
C  =  119°22'27i", 


c'=162°34'27",      C'=164°41'56". 


322 


SPHERICAL   THIGONOMETRY. 


■  i 

1  , 

1 

70. 

Given  A  =  52°  50' 20", 

B=  66°   7' 20", 

«=    59°  28' 27";          v; 

:           1 

find    b=  81°  15' 15", 

c=  110°  10' 50^", 

0=119°  43' 48", 

ipf^ 

71. 

or      b'=  98°  44' 45", 
Given  A =115°  36' 45", 

c'=138°45'26", 
B=  80°  19' 12", 

C'=142°24'59". 
b=   84°21'56"; 

.  ■  '    ■  1 

find   a =114°  26' 50", 

c=  82°  33' 31", 

C=  79°  10' 30". 

I     ■                    72. 

Given  A  =  61°  37' 52". 7, 

B  =  139°54'34".4, 

6=150°17'26".2; 

1 

find   a=  42°37'17".5, 

0=129° 41'  4".8, 

C=  89°54'19".0, 

■v  .                          .        ^ 

or      a'=137°22'42".5. 

c'=  19°58'35".6, 

Q/1 

C'=  26°21'17".6. 

_Jh:                             iJ.   iS 

|!                   73. 

Given  A  =  70°, 

B  =  120°, 

&=  80°. 

:9H|lBn 

Ans.  Impossible  ;  why  ? 

cS7 

' '  ^^HHI 

1                   74. 

Given  a=  108°  14', 

b=  75°  29', 

c=  56°  37'; 

! 

find  A  =  123°  53' 47", 

B=  57°  46' 56", 

C=  46°51'51".5.     ^"'^• 

'In 

1                    75. 

Given  a=  57°  17', 

b=  20°  39', 

c=  76°  22'; 

find  A=  21°  1'  2", 

B=     8°  38' 46", 

C  =  155°31'36".5.     ^^■ 

■■fe^^^^  ,2-^„LjM 

76. 

Given  a=  68°  45', 

b=  53°  15', 

c=  46°  30'; 

find  A=   94° 52' 40", 

B=  58°  5' 10", 

C=  50°50'52|".       c) 

H 

1      •              77. 

Given  a=   63°  54', 

&=  47°  18', 

c=  53°  26'; 

1  " 

find  A=  86°  30' 40", 

B=  54°  46' 14", 

C=  63°12'55i". 

78. 

Given  a=  70°  14' 20", 

6=  49°24'10", 

c=   38°  46' 10";        0 

! 

find  A= 110°  51 '16", 

B=  48°  56'. .4", 

C=  38°26'48". 

9 
c=  97°12'25"; 

79. 

Given  a=124°12'31". 

b=  54°  18' 16", 

''^^HII 

find  A=127°22'   7", 

B=  51°  18' 11", 

C=   72°  26' 40". 

H|                     80. 

Given  a=  60°  12'  4", 

?>=116°44'48", 

c=129°ll'42"; 

find  A=  59°  4' 26", 

B=  94°  23' 10", 

C  =  120°  4' 50". 

HI        ^^' 

Given  a  =100°, 

b=  50°, 

c=   60°;                 ),*j 

find  A=138°15'46".4, 

'b=  31°11'14".0, 

C=  35°49'68".2. 

^H                 ^^' 

Given  A  =  86°  20', 

B=   76°  30', 

C=  94°  40';           ,'^] 

1 

find  a«  87° 20' 28", 

bm  76''44'2i", 

c«  93°  66' 31".     lat 

EXAMPLES. 


59°  28' 27"-, 
19°  43' 48", 
42°  24' 59". 

84°  21' 56"; 
79°  10' 30". 

L50°17'26".2; 
89°54'19".0, 
26°21'17".6. 

80°. 

56°  37'; 
46°51'51".5. 

76°  22'; 
:155°31'36".5. 

:  46°  30'; 

:  50°50'52i". 

:  53°  26'; 

=  63°  12' 551". 

=  38°  46' 10"; 
=  38°  26' 48". 

=  97° 12' 25"; 

=  72°  26' 40". 

=  129°  11' 42"; 
=  120°  4' 50". 

=  60°; 

=  35°49'58".2, 

=  94°  40'; 
=3  93°  56' 31". 


83.!  Given  A  =  96°  45', 

find   a=  88°  27' 49", 

184.  Given  A  =  78°  30', 
find   a=  74° 57 '46", 

1 85.  Given- A  =  57°  50', 
find   a=  58°  8' 19", 

1 86.  Given  A  =  129°  5' 28", 
find   a =135°  49' 20", 

|87.   Given  A  =  138°  15' 50", 
find   a=100°  0'  8". 4, 

|88.   Given  A= 102°  14' 12", 
find   a=104°25'  8", 

89.   Given  A  =  20°  9' 56", 
find   a=  20°  16' 38", 


B  =  108°30', 
6=107°  19'52", 

B  =  118°40', 
6=120°  8' 49", 

B=  98°  20', 
b=  83°   5' 36", 

B  =  142°  12' 42", 
6 =144°  37 '15", 

B=  31°  11' 10", 
6=  49°59'56".4, 

B=  54°  32' 24", 
b=  53°  49' 25", 

B=  55°  52' 32", 
b=  56°19'41", 


323 

C=116°15'; 
c=115°28'13V'. 

C=  93°  20'; 
c=100°18'llf". 

C=  63°  40'; 
c=  64°  3' 20". 

C  =  105°  8' 10"; 
c=   60°  4' 54". 

C=  35°  50'; 
c=   60°  0'11".2. 

C=  89°  5' 46"; 
c"=  97°  44' 18". 

C  =  114°  20' 14"; 
c=  66°  20' 43". 


^i 


90. 


If  a,  b,  c  are  each  <  ^,  show  that  the  greater  angle  may 


txceed  -• 
2 

91.  If  a  alone  >^7r,  show  that  A  must  exceed  ^• 

92.  If  a  and  b  are  each  >^7r,  and  c  <  ^tt,  prove  that : 

(1)  The  greatest  angle  A  must  be  >i'n-; 

(2)  B  may  be  >  j^ir; 

(3)  C  may  or  may  not  be  <  ^tt. 

93.  If  cos  a,  cos  b,  cos  c  are  all  negative,  prove  that  cos  A, 
)s  B,  cos  C  are  all  necessarily  negative. 

94.  In  a  spherical  triangle,  of  the  five  products,  cos  a  cos  A, 
IS  h  cos  B,  cos  c  cos  C,  cos  a  cos  b  cos  c,  ~  cos  A  cos  B  cos  C,  show 
lat  one  is  negative,  the  other  four  being  positive. 


I 


] 


324 


SPHERICAL    TRIGONOMETR  F. 


A 


■  I  ■ 


CHAPTER   XII. 

THE  IN-OIROLES  AND  EX-OIEOLES.  —  AREAS. 

215.  The    In-Circle    (Inscribed    Circle). —  To  find    the 
angular  radius  of  the  in-circle  of  a  triangle. 

Let  ABC  be  the  triangle;  bisect  the  angles  A  and  B 
by  the  arcs  AG,  BO;  from  0  draw 
OD,  OE,  OF  perpendicular  to  the 
sides.  Then  it  may  be  shown  that 
0  is  the  in-centre^  and  that  the  per- 
pendiculars OD,  OE,  OF  are  each 
equal  to  the  required  angular  radius. 

Let  2  s  =  the  sum  of  the  sides  of 
the  triangle  ABC.  The  right  triangles 
OAE,  OAF  are  equal. 

.-.  AF  =  AE. 

Similarly,  BD  =  BF,  and  CD  =  CE. 

.-.  BC  +  AF  =  AC  +  BF  =  s. 
.-.  AF  =  s-BC  =  8-a. 
Now  tan  OF  =  tan  OAF  sin  AF    .     (Art.  186) 

or,  denoting  the  radius  OF  by  r,  we  have 


tan  r  =  tan  —  sin  (s  —  a) 
2        ^  ^ 


(1) 


or     tan  r 


Vsin  (s  —  g)  sin  {s  —  b)  sin  {s  —  c) 
sins 


n 


sins 


(Art.  195)  (2) 


THE  ESCEIBED  CIRCLES. 


325 


a) 


Also,  sin(s— a) 

=sin ^(6  +  c)  cos  -|-a  —  cos  ^(6  +  c)  sin  ^a 
sin  ^  a  cos  ^  a 


.    A 
sin  — 

2 


[cos  |(B  -  C)  -  cos  i(B  +  C)]      (Art.  198) 


_  sinct  sin^B  sin^^C 
sin^A 

which  in  (1)  gives 

.    B    .   -C 

sin  —  sill  - 
,                   2        2. 
tan  r  = :— : —  sm  a 


cos  ^  A 


N 


,    .    .    .     (3) 
(Art.  196)  (4) 


2  cos  ^  A  cos  ^B  cos|-C 
an  equation  which  is  equivalent  to  the  following: 

cot7'  =  ^— [cosS+cos(S-A)+cos(S-B)+cos(S-C)](5) 

^  JN 

216.  The  Ex-Circles.  —  To  find  the  dtngular  radii  of  the 
ex-circles  of  a  triangle. 

A  circle  which  touches  one  side  of  a  triangle  and  the 
other  two  sides  produced,  is  called  an  escribed  circle,  or 
ex-circle,  of  the  triangle.  It  is  clear  that  the  three  rx-circles 
of  any  triangle  are  the  in-circles  of  its  colunar  triangles 
(Art.  191,  Sch.). 

Since  the  circle  escribed  to  the  side  a  of  the  triangle 
ABC  is  the  in-circle  of  the  colunar  triangle  A'BC,  the  parts 
of  which  are  a,  ir  —  b,  tt  —  c, 
A,  TT  —  B,  IT  —  C,  the  problem 

becomes   identical  with   thiit   A<^        I  Y  ^^A 

of  Art.  215 ;   and  we  obtain 
the  value  for  the  in-radins  of 
the  colunar  triangle  A'BC,  by  substituting  for  b,  c,  B,  C, 
their  supplements  in  the  five  equations  of  that  article. 


JiT    ' 


326  SPHERICAL   TEIGONOMETJiY. 

Hence,  denoting  the  radius  by  i\,  we  get 

tan  ?•„  =  tan  I A  sin  s (1) 

=^- — (2) 

sm(s  — a) 

cos 4  B  cos 4  C   ■  /o\ 

= 2 2 —  gin  a (3) 

cos  \  A 

^ NL (4) 

2cos^Asin^Bsin^C     '     '■   '    '     ^  ' 

cotr,=--[-cosS-cos(S-A)+cos(S-B)+cos(S-C)](5) 

These  formulae  may  also  be  found  independently  by 
methods  similar  to  those  employed  in  Art.  215,  for  the 
in-circle,  as  the  student  may  show. 

Sch.  Similarly,  another  triangle  may  be  formed  by  pro- 
ducing BC,  BA  to  meet  again,  and  another  by  producing 
CA,  CB  to  meet  again.  The  colunar  triangles  on  the  sides 
b  and  c  have  each  two  parts,  b  and  B,  c  and  C,  equal  to 
parts  of  the  primitive  triangle,  while  their  remaining  parts 
are  the  supplements  in  the  former  case  of  a,  c.  A,  C,  and  in 
the  latter,  of  a,  b,  A,  B. 

The  values  for  the  radii  r^  and  r<.  are  therefore  found  in 
the  same  way  as  the  above  values  for  r„ ;  or  they  may  be 
obtained  from  the  values  of  ?•„  by  advancing  the  letters. 

Thus,  tan  Vi,  =  tan  I B  sin  s  = ,  etc., 

sin(.s  — 6) 

and  tan  r^  =  tan  i  C  sin  s  =  -r-- ,  etc. 

sin  (s—c) 

217.  The  Circumcirde.  —  To  find  the  angular  radius  of 
the  circumcirde  of  a  tnanyle. 

The  small  circle  passing  through  the  vertices  of  a  spheri- 
cal triangle  is  called  the  circumscribing  circle,  or  circumcirde, 
of  the  triangle. 


(1) 

(2) 
(3) 


.    .     (4) 

-C)]  (5) 

ently  by 
,  for  the 

d  by  pro- 
producing 
the  sides 
equal  to 
ling  parts 
C,  and  in 


found  in 

)y  may  be 
etters. 

itC., 

tc. 

•  radius  of 

»f  a  spheri- 
ircumcircle, 


THE  CIRCUMCIRCLE. 


327 


Let  ABC  be  the  triangle;  bisect  the 
sides  CB,  CA  at  D,  E,  and  let  0  be  the 
intersection  of  perpendiculars  to  CB, 
CA,  at  D,  E;  then  0  is  the  circuiu- 
centre. 

For,  join  OA,  OB,  OC;  then  (Art.  186) 

cos  OB  =  cos  BD  cos  OD, 
cos  OC  =  cos  DC  cos  OD. 
.-.  OB  =  OC.     Similarly,  OC  =  OA. 
Now  the  angle 

OAB  =  OBA,  OBC  =  OCB,  OCA  =  OAC. 
.-.  OCB  +  A=|(A  +  B  +  C)=S. 

.-.  OCB  =  S-A. 
Let  OC  =  R ;  then,  in  the  triangle  ODC,  we  have 

cos  OCD  =  tan  CD  cot  CO  =  tan  |  a  cot  R  .     (Art.  18G) 

tan  4  a 


.-.  tan  R  = 


or 


tan  R  =  — 


cos(S- A) 
cosS 


N 


.    .    .     .     (1) 
(Art.  196)  (2) 


Also  cos(S  -  A)=  cosi[(B  +  C)  -  A] 

=  cos|(B  +  C)  cos|A  +  sin|(B  +  C)  sin|A 

=  sin^^?iA[-cos|(64.c)4.cosK&-c)]  (Art.  198) 

COS  ■«-  ct 
sin  A 


cos^a 


cos^&cos|c, 


which  in  (1)  gives 
tanR  = 


sin^g 


(3) 


sin  A  cos  1 6  cos  ^c 

2sin4a  sin46sin4c         ,.   ,    ^..j,,    ,.. 
= ^ 2 2_  .     (Art.  195)  (4) 


II 


i! 


£ 


i:- 


M; 


f , 


&28  BPHEBICAL   TttlGONOMETHY. 

which  may  be  reduced  to  the  following : 

tan  R  =  — [8in(s  —  a)  +  sin(s  —  h)-\-  sin(s  —  c)  —  sin  s]   (5) 
2n 

218.  Circumcircles  of  Colunar  Triangles  —  To  find  the 
angular  radii  of  the  circumcircles  of  the  three  colunar  triangles. 

Let  Rj,  Rgj  ^3  ^6  the  angular  radii  of  the  circumcircles  of 
the  colunar  triangles  on  the  sides  a,  6,  c,  respectively. 
Then,  since  Ri  is  the  circumradius  of  the  triangle  A'BC 
whose  parts  are  a,  v  —  b,  tt  —  c,  A,  tt  —  B,  tt  —  C,  we  have, 
from  Art.  217, 

tanRi  =  -*?5i^ (1) 

cosS 

tanR,=:"^^(^-^) (2) 

tan  Ri  = -— -4^^i^^-- (3) 

sin  Asin^o  sm^c 

tan  R,  =  ^  ^"^  ^  g  cos  ^  6  cos  ^  c ^     ,^. 

tan  Rj  =  — -  [sins— sin(s— a)+sin(s— 6)+sin(s— c)]  (6) 

Similarly, 

tan  R,  =  -  *^5Li^  =  ^^^i^Jni^  =  etc., 
cosS  N  ' 

and  tan  R,  =  -  ^-^^^  =  ^i^-Cl  ^ 
'  cos  S  N 

EXAMPLES. 
Prove  the  following : 

1.  cos  8  4-  cos  (s  —  a)  +  cos  {s  —  b)-\-  cos  (s  —  c) 

=  4  cos  ^  a  cos  ^  6  cos  ^  c. 

2.  cos(s  —  6)-f  cos(s  —  c)  —  cos(8  —  a)  — COSS 

=  4  cos  ^  a  sin  ^  b  sin  J^  c. 


^ 


PROBLEM. 


329 


3.  tann  =  52iiCc^si„^,^ N 

cos^B  2  cos  ^  13  sin  ^C  sin  ^  A 

4.  tanr,=.^^^iA^iMsinc  =  — — ^^^ 

cosiC  2  cos  I C  sin  ^  A  sin  ^B* 

6.   cot  r  :  cot  Vi :  cot  i-g :  cot  r^ 

=  sin  s  :  sin  (s  -  a) :  sin (s  -  b)  :  sin (s  -  c). 

6.  tan  r  tan  r^  tan  ?-2  tan  r^  =  nl 

7.  cot  r  tan  rj  tan  ?'a  tan  r^  =  sin^s. 

8.  tanR2  =  2£2^i5L^li£2i±f. 

n 

9.  tanR3  =  ^cosiacosi6^inic_ 

10.  tanRi:tanR2:tanR3r^cos(S-A):cos(S-B):cos(S-C). 

11.  cot  R  cot  Rj  cot  Eg  cot  R3  =  N". 

12.  tan  R  cot  Rj  cot  R2  cot  R3  =  cos"  S. 


i 


AREAS    OF    TRIANGLES. 

219.  Problem.  —  Tojind  the  area  of  a  spherical  triangle, 
having  given  the  three  angles. 

Let   r  =  the  radius  of  the  sphere. 

E  =  the  spherical  excess  =  A  +  B  -f  C  —  180°. 
K  =  area  of  triangle  ABC. 

j-t  is  shown  in  Geometry  (Art.  738)  that  the  absolute  area 
of  a  spherical  triangle  is  to  that  of  the  surface  of  the  sphere 
as  its  spherical  excess,  in  degrees,  is  to  720°. 


.-.  K:47rt-2=E:720°. 

...  K  =  -^-,rr2 
180° 


•  • 


(1) 


330 


SPUEliWAL   TRIGONOMETRY. 


Cor.   The  areas  of  the  colunar  triangles  are 

(^2A-E)     ,    1215- E),    (2C_-E)^ 
180°      ^'        180°      "  '        180° 


i!       4 


220.  Problem.  —  To  jind  the  area  of  a  tnangle,  having 
given  the  three  sides. 

Here  the  object  is  to  express  E  in  terms  of  the  sides. 
I.    CagnolVs  Theorem. 
sin  IE  =  sin  ^(A  +  B  +  C  -  tt) 

=  sin  ^(A  +  B)  sin  ^C  —  cos  ^(A  +  B)  cos  ^C 

=  5^5LK^^-[oosi(a-?/)-cosi(«  +  6)]    •  (Art.  198) 
cos^c 

_  sin  ^  g  sin  ^6  sin  C  _  sin  ^  a  sin  ^b  2n        /  a  .f  1Q''\/'1N 

cos  ^ c  cos  ^ c        sin  «  sin  6 


.-.  sin^E  = ''^ 

2  cos  ^a  cos  ^6  cos  ^c 

II.   Lhuilier^s  Theorem. 
^         cosKA+B  +  C-tt) 


(2) 


^sini(A  +  B)-sinK^-C)  (j^j.,  ak>. 

cos^(A4-B)  +  cos|(7r-C) ^       '      ^ 

_sin|(A  +  B)  — cos|C 
~  cos  ^( A  +  B)  +  sin  I C 


cos  ^{a  —  b)  —  cos  ^c    cos^C 
cos  ^(a +  6)  + cos  ^c    sin^C 


,     .    .     .  (Art.  li)6) 

.    .     .     .     .     (Art.  4.5) 
=  Vtan|.stan|(s-a)tan^(.s-6)tan|(s-c)    (Art.  195)  (3) 


^  sin  i(s  -  b)  sin  ^(s  -  a)  ^^^  ^  ^ 

COSi^SCOS^(.S  — c) 


ASEAS  OF  TlilANOLES. 


331 


221.  Problem.  —  To  find  the  area  of  a  triangle,  having 
given  two  sides  and  the  included  angle. 

co8^E=  cos  [^  (A  -f-  B)  -  (^TT  -  :|C)] 

=  cos^( A  +  B)  sin^C  +  sin^(A  -|-  B)  cos^C 

=  co8^{a  +  b)fim^C  +  cosi{a-b)cm^C    (Art.  19<S) 

=  [cos|acos^?>  +  sin|asin^&cosG]sec^c    .     .     (1) 

Dividing  (1)  of  Art.  220  by  this  equation,  and  reducing, 
we  have 

tanjE-     tanjatan^fesinC        

1  +  tan  4  a  tan  ^  h  cos  C  '    '     '     \  ) 


EXAMPLES. 

1.  Given  a  =  113°  2'  56".64,  b  =  82°  39'  28".4,  c  =  74°  54' 
31  ".06;  find  the  area  of  the  triangle,  the  radius  of  the 
sphere  being  r. 

By  formula  (3)  of  Art.  220, 

a  =113°  2'56".64 


6=    82°39'28".40 
c=    74°  54' 31  ".06 

2s  =  270°36'56".10 


s  =  135°18'28".05 

a^.   22°15'31".41, 
b=    52°38'59".65, 
s-c=   60°23'56".99. 


8 
8 


K = mm  X  TT^-^ . 

=  IMMf»''  =  area. 


is  =  67°39'14".025 
^(.s--a)  =  ll°  7'45".705 
i(s-6)  =  26°19'29".825 
|(s-c)  =  30°ll'58".495 

logtan  ^8  =  0  3860840 
log  tan  ^(.s  -  a)  =  y.2938583 
log  tan  l(s  -b)  =  9.6944058 
log  tan  i  (s  -  c)  =  9. 7649261 

logtan2^E  =  9.1392742 

log  tan  ^E  =  9.5696371. 

iE  =  20°21'58".25. 
E  =  81°  27' 53" 
=  293273". 

.     .     [(1)  of  Art.  219] 


ill 


^i!  :■! 


» 


832 


SPHERICAL   TRIGONOMETRY. 


2.  Given  A  =  84°  20' 19",  13  =  27°  22' 40",  C  =  75°33'; 
find  E  =  7°15'59". 

3.  Given  a  =  40°  24',  6  =  07°  14',  c  =  81°12'; 
find  K  = 


4.  Given  a  =108°  14',  6  =  75°  29',  c  =  56°37'; 
find  E  =  48°  32'  34".5. 

5,  Prove  cos4E  =  1  +  ""^"  +  ^^^^-  +  ""^^ 

4  cos  ^  a  cos  ^  6  cos  ^  c 

__  cos'^g  +  cos'^6  -f-  cos^^c  —  1_ 
2cos^acos^6cos|c 


7. 


6.       «  ainiE=vP^^^"^^^~"^^"^^^^~^)^"^^(^~^) 
\  cosia  cosi^cosAc. 


cos^a  cos  1^6  cos^^c. 


«  C03lE=vfc^^^^"^^^^~^^^"^^^^~^^^"^^^^~^^ 

\  cos^acos^ftcos^c 


8.       "  eottE^:^"^^^^"^^^'^^^^^ 
^  sin  C 

_eot|6cot^c  +  C0S  A 
sin  A 

cot^ccot^g-t-cosB 
sinB 


EXAMPLES. 
Prove  the  following : 

1.  sin  (s  —  a)  +  sin  (s  —  &)  +  sin  (s  —  c)  —  sin  s 

=  4  sin ^ a  sin ^6  sin^c. 

2.  sin  s  +  sin  (s  —  6)  +  sin (s  —  c)  —  sin  (s  —  a) 

=  4sin^a  cos^6  cos^c. 


EXAMPLES. 


333 


3.  sin(s  — 6)sin(.s  —  o) -f  siii(.'}  —  c)8iu(s  — a) 

+  sin  (a  —  a)8in(s  —  6)  +  sins  sin  (s  —  a) 
+  sill  s  sin  (.9  —  b)-\-  sin  8  sin  (a  —  c) 
=  sin  b  sin  c  +  sin  c  sin  a  -f  sin  a  sin  6. 

4.  sin(.«»  —  6)sih(s  —  c)  +  sin(s  —  c)sin(.s  —  a) 

—  sin(s  —  a)sin(.s  —  6)  +  sins  8in(s— (;) 
4-  sin  8  yin  (.s  —  b)~  sin  s  sin  (.s  —  c) 
=  sin  6  sin  c  +  sin  c  sin  a  —  sin  a  sin  b. 

5.  sin's  4-  8in*(s  —  «)  +  sin-(s  —  6)  +  8in^(s  ~  c) 

=  2(1  —  cos  rt  cos  &  cose). 

6.  sin's  +  sin''(s  —  a)  —  sin'(8  —  6)  —  sin'(s  —  c) 

=  2  cos  a  sin  6  sine. 

7.  cos's  4-  cos'(s  —  a)  +  cos'(s  —  &)  -f  cos'(s  —  e) 

=  2(1  -}-  cos  a  cos  &  cose). 

8.  cos's  +  cos'(s  —  a)  —  cos'(«  —  6)  —  cos'(s  —  c) 

=  —  2  cos  a  sin  6  sin  e. 

9.  tan  r  cot  Vi  tan  rj  tan  rg  =  sin'  ( s  —  a) . 

10.  tan  r  tan  Vi  cot  rj  tan  rg  =  sin'  (s  —  6) . 

11.  tan r  tan  rj  tan  j'a  cot  9-3  =  sin'(s  —  c) . 

12.  cotr  sins  =  cot^Acot^B  cot^C. 


13.   tan  Vi  +  tan  r^  -f-  tan  r^  —  tan  r  = 


4NsinS 


sin  A  sin  B  sin  C  '" 


Hj         .       .       ,       .       .            ■           4sin4^asini^6sin4c 
14.   cot  r,  +  cot  r^  +  cot  ?  3  —  cot  r  = ^ — ^ — 


15.   tan  r,  :  tan  n  :  tan  r. 


sma 


sin  & 


sin  c 


1  4-  cos  A    1  4-  cos  B    1  +  cos  C 


^e     tanr,+tanr2  4-tan  rs— tan  7-      i,^  •  „^„„  •  „^„i.  .  „..„„\ 

16.    — ~ =  *(14-cosa4-coso4-cosc). 

cot  Vi  4-  cot  rj  4-  cot  r^ — cot  r 


:r]^ 


■:1 


334 


SPHERICAL   TRIGONOMETRY. 


n' 


17.    cotV.  +  cot'^r,  +  cot=r3  +  cot^r  =  2(l-cosacqs6^osc). 

18. 
19. 


_J_,_^ 1 !_ 

sin'^r     siirri     siu-rj     siii^rs 


2  cos  g  sin  6  sin  c 
»2 


cot  r-i  cot  '/'g  +  cot  j'a  cot  rj  +  cot  i\  cot  7', 

+  cot  r(cot  r,  +  cot  rg  +  cot  r^ 

_  sin  b  sin  c  +  sin  o  sin  a  +  sin  a  sin  h 
_  _ 

20.  tan  »*2  tan  r^  +  tan  r^  tan  j'l  -f  tan  r,  tan  rj 

4-  tan  r(tan  )\  +  tan  ro  -f  tan  r^) 
=  sin  b  sin  c  +  sin  c  sin  a  +  sin  a  sin  6. 

21.  cot  R  tan  R,  cot  Rg  ^ot  R.,  =  cos-  (S  —  A) . 

22.  cot  R  cot  Ri  tan  R^  cot  Kg  =  cos-  (S  -  li). 

23.  cot  R  cot  Ri  cot  Rg  tan  Rg  =  cos^  ( S  -  C) . 

24.  tan  Ri  -f  tan  Rg  =  cot  r  +  cot  r^. 

25.  tan  R,  +  tan  Rj  -f  tan  Rg  —  tan  R  =  2  cot  r. 

26.  tan  II  —  tan  lli  +  tan  R,  +  tan  Rg  =  2  cot  i\. 

27.  tan  R  +  tan  Rj  —  tan  Rg  -|-  tan  Rg  =  2  cot  r^ 

28.  tan  R  +  tan  R,  +  tan  Rj  —  tan  Rg  =  2  cot  r^. 

29.  cot  Vy  +  cot  Ti  4-  cot  rg  —  cot  r  =  2  tan  R. 

30.  cot  r  —  cot  )\  +  cot  r^  +  cot  rg  =  2  tan  Rj. 

31.  cot  )•  4-  cot  r,  —  cot  r^  -|-  cot  rg  =  2  tan  Rj. 

32.  cot  r  +  cot  ri  4-  cot  rj  —  cot  r^  =  2  tan  R,. 

33.  tan  R  4-  cot  r  =  tan  R,  4-  cot  r,  =  etc., 

=  \  (cot  r  4-  cot  r,  4-  cot  rj  4-  cot  i\). 

34.  tan^  R  4-  tan*  R,  4-  tan'^  Kj  4-  tan-  R., 

2(1  4-<'os  A  cosB  cosC) 


T?'^ 


35. 
36. 


EXAMPLES.  335 

tan'  R  +  tan'  R,  +  tan"  R^  +  tan^  R,  _  . 
cot^-  +  cot2ri  +  cot=*?-2  +  cot'r3     ~    * 

tan'  R  +  tan'  R^  -  tan'  B^  -  tan'  R^ 

_     2(cos  AsinBsinC) 


37. 

38. 

39. 
40. 

41. 


cot'  r  +  cot'  I'l  —  cot'  7-2  —  cot'  Va  = 


N' 
2  cos  a  sin  b  sin  c 


ji^" 


tan'  R  +  tan'  R,  -  tan'  Rg  -  tan' R.,  _  _  cos  A 
cot' r  +  cot'  Vi  —  cot' rg  -  cot'  r^     ~      cos  a* 

tan  R  cot  R,  =  tan  ^ 6  tan  ^c. 

(cot r  4-  tan R)'-f  1  =  /'»i"Cf'  + si"^>  +  sincV 

(cot  r,  -  tan  R)'+  1  =  A'^"  ^  +  sin  o  -  sin  gV^ 


42.    tan  J  A  sin  (s  — a)  = 


2  cos  ^  A  cos  ^B  cos  \Q 


43. 
44. 

45. 

46. 

47. 

48. 


tan  r  _  cos(S  -  A)  cos(S  -  B)  cos(S  -  C) 
tanR  2  cos  ^  A  cos  ^B  cos  ^C 

cot(s  —  6)  cot(s  —  c)  +  coL(.s  —  r')cot(.s  —  a) 

+  cot  (.9  —  a)  cot(s  —  6)  =  cosec'r. 
cot(8-6)cot(s— c)  -  cotscot(s— 6)  —  cot8cot(s-c) 

=  cosec'r,- 

cot(s— c)cot(«— a)  —  cot«cot(s— c)  —  cot«cot(«— a) 

=  cosec'r2. 

cot(s— a)cot(«— 6)  —  cot  5  cot  (s— a)  —  cots  cot  (s  —  6) 

=  cosec'^7'3. 

cot(s  — g)      cot(jt  — 6)      cot(«  — c)      2  cot  a 
sin'r,  sin'rj  sin'r,  sin'r 

"s  3cot(»  —  a)  cot(8  —  6)cot(5  —  c). 


336 


8PHERWAL    TRIGONOMETRY. 


Wl-    '■  Si     V 


49. "  cosec'ri  +  cosec'/'a  +  cosec';'3  —  cosec'r 

=  —  2  cot  s[cot(s  —  a)  +  cot(s  —  &)  +  cot(s  —  c)]. 


50. 


1 


+^+^ 


+  - 


sin^r     sin^ri     sin'^7*2     sin''?  3 


51.   cot  R  —  cot  El  —  cot  R2  —  cot R3  = 


—  2  S  tan  {s  —  ft)  tan  {s  —  h) 
tan  s  tan  (s  —  a)  tan  (s  —  b)  tan  (s  —  c) 

2  N  cos  8 


n 


52.    sin(A-iE)  =  - 


n 


2 cos  ^a sin  ^ftsin^c 


53.    sin(B-iE)  =  - 


n 


54.    sin(C-^E) 


2sin^acos^6sini^c 

n 

2  sin ^  a  sin  ^6  cos  ^c 


55.    cos(A-:^E)  =  — --5— -i; — .^.  ,    .    /    ♦ 

2co8^ttsin^o  sin  J^c 


56.    cos(B-iE) 


sin'^c  +  sin'^^ft—  sin^^ft 
2sin^acos^6sin  ^c 


57.  co8(C  -  i  E)  =  '"^^"'^^  +  ^^"'i^  -  ^"^H^. 

2  sin  1^^ a  sin  ^  &cos^c 

58.  cot(A-^E)  =  "^H^tani&-cosC 

^         ^     ^  sinC 

_  tan  ^  6  tan  ^  c  +  cos  A  _  tan  ^  a  cot  1 6  —  cos  B 
sin  A  sin  B 


59.   tan^(A— ^E)  =  Vcot^scotJ^(s— a)tan^(8— 6)tan^(s— c). 


60.    tan^(B  — .J^E)  =  Vcotistan^(8— a)cot|(,s— 6)tanJ^(.s— c). 


61.   tan^(C— J^E)=  Vcot^«tan|(«— a)tan|(s— 6)cot^(s— c). 

62.  If  S,  Si,  Sj,  S^denotc  the  sums  of  the  angles  of  a  triangle 
and  its  three  colunars,  prove  that  S  +  S,  +  Sj  +  Sg  =  3  tt. 

63.  In  an  equilateral  triangle,  tan  R  =  2  tanr. 


EX  A  HPLES. 


337 


64.  If  El,  Ej,  Eg  denote  the  spherical  excesses  of  the  colunars 
on  a,  h,  c,  respectively,  show  that  E  +  Ei  +  Eg  +  E3  =  27r ; 
and  therefore  the  sum  of  the  areas  of  any  triangle  and  its 
colunars  is  half  the  area  of  the  sphere. 

66.   Given  a  =  108°  14',  6  =  75°  29',  c  =  56°37'; 
find  E  =  48°  32'  34".5. 

66.  Given  a  =  63°  54',  &  =  47°18',  c  =  53°26'; 

find  E  =  24°  29'  49^". 

67.  Given  a  =  69°  15' 6",  6  =  120°  42' 47",  c  =  159°  18' 33"; 

find  E  =  216°  40' 23". 

68.  Given  a  =  33°i'45",  &  =  155°  5' 18",  C  =  110°10'; 

find  E  =  133°  48'  55". 

69.  Given  a=.l=c  =  1",  on  the  earth's  surface ; 

find  E  =  27".21. 

70.  Given  a  =  b  =  c  =  60°,  on  a  sphere  of  6  inches  radius ; 
find  the  area  of  the  triangle.         Ans.  19.845  square  inches. 

71.  If  a=6=^,  and  c=^,  prove  sin^E=^,  and  cosE=|. 

72.  If  C  =  ^,   prove   sin  ^  E  =  sin  ^  a  sin  ^  6  sec  ^  c,   and 

ill 

cos^^E  =  cos^acos^fi  sec^c. 

73.  If  a  =  6,  and  C  =  -,  prove  tan  E  =  ^  tan  a  sec  a. 

74.  If  A4-B4-C  =  27r,  prove  cos''^a4-cos''^6  +  cos2^c=l. 

75.  If  a  +  6  =  TT,  prove  that  E  =  C  ;  and  if  E'  denote  the 
spherical  excess  of  the  polar  triangle,  prove  that 

sin^E'  =  sinaco8^C. 


;  37r. 


nc    -o  -211?      VsmTEsinlEiSinAEsSin^Ea 

76.  Prove  8mHE  = '    ^ .        / , .,  W ' 

cot  \  a  cot  ^  0  cot  i  c 


838 


SPHERICAL   TRIG  ON  O  METR  Y. 


Uy\ 


CHAPTER   XIII. 


APPLIOATIONS   OF  SPHEEIOAL  TRIGONOMETEY. 


SPHERICAL   ASTRONOMY. 


■t  i 


* 


222.  Astronomical  Definitions, 

The  celestial  S2)here  is  the  imaginary  concave  surface  of 
the  visible  lieavens  in  which  all  the  heavenly  bodies  appear 
to  be  situated. 

The  sensible  horizon  of  a  place  is  the  circle  in  which  a 
plane  tangent  to  the  earth's  surface  at  the  place  meets 
the  celestial  sphere. 

The  rational  horizon  is  the  great  circle  in  which  a  plane 
through  the  centre  of  the  earth  parallel  to  the  sensible 
horizon  meets  the  celestial  sphere.  I^ecause  the  radius 
of  the  celestial  sphere  is  so  great,  in  comparison  with  the 
radius  of  the  earth,  these  two  horizons  will  sensibly 
coincide,  and  form  a  great  circle  called  the  celestial  hori- 
zon. 

The  zenith  of  a  place  is  that  pole  of  the  horizon  which 
is  exactly  overhead ;  the  other  pole  of  the  horizon  directly 
ur.derneath  is  called  the  nadir. 

Vertical  circles  are  great  circles  passing  through  the 
zenith  and  nadir.  The  two  principal  vertical  circles  are 
the  celestial  meridian  and  the  prime  vertical. 

The  celestial  meridian  of  a  place  is  the  great  circle  in 
which  the  plane  of  the  terrestrial  meridian  meets  the 
celestial  sphere ;  the  points  in  which  it  cuts  the  horizon 
are  called  the  north  and  south  points. 


SPHERICAL  ASTRONOMY. 


339 


The  prime  vertical  is  the  vertical  circle  which  is  per- 
pendicular to  the  meridian ;  the  points  in  which  it  cuts 
the  horizon  are  called  the  east  and  ivest  points. 

The  axis  of  the  earth  or  of  the  celestial  sphere  is  the 
imaginary  line  about  which  the  earth  rotates. 

The  celestial  equator,  or  equinoctial,  is  the  great  circle  in 
which  the  plane  of  the  earth's  equator  intersects  the 
celestial  sphere. 

The  poles  of  the  equinoctial  are  the  points  in  which  the 
axis  pierces  the  celestial  sphere. 

Hour  circles,  or  circles  of  declinathn,  are  great  circles 
passing  through  the  poles  of  the  equinoctial. 

The  ecliptic  is  a  great  circle  of  the  celestial  sphere,  and 
the  apparent  path  of  the  sun  due  to  the  real  motion  of 
the  earth  round  the  sun. 

The  equinoxes  are  the  points  in  which  the  ecliptic  cuts 
the  equinoctial.  There  are  two,  called  the  vernal  and  the 
autumnal  equinox,  which  the  sun  passes  on  March  20  and 
September  22. 

The  obliquity  of  the  ecliptic  is  the  angle  between  the 
planes  of  the  ecliptic  and  equator,  and  is  about  23°  27'. 

Circles  of  latitude  are  great  circles  passing  through  the 
poles  of  the  ecliptic. 


which 
irectly 


223.  Spherical  Coordinates.  —  The  position  of  a  point 
on  the  celestial  sphere  may  be  denoted  by  any  one  of  three 
systems.  In  each  system  two  great  circles  are  taken  as 
standards  of  reference,  and  the  point  is  determined  by 
means  of  these  circles,  which  are  called  its  spherical, 
coordinates,  as  follows: 


I.    The  horizon  and  the  celestial  meridian  of  the  place. 

The  azimuth  of  a  star  is  the  arc  of  the  horizon  inter- 
cepted between  the  south  point  and  the  vertical  circle 


if 


^t  > 


i& 


340 


SPHERICAL   TRIGONOMETRY. 


passing  through  the  star;  it  is  generally  reckoned  from 
the  south  point  of  the  horizon  round  by  the  west,  from 
0"  to  360". 

The  altitude  of  a  star  is  its  angular  distance  above  the 
horizon,  measured  on  a  vertical  circle.  The  complement 
of  the  altitude  is  called  the  zenith  distance. 

II.  Tlie  equinoctial  and  the  hour  circle  through  the  vernal 
equinox. 

The  right  ascension  of  a  star  is  the  arc  of  the  equinoctial 
included  between  the  vernal  equinox  and  the  hour  circle 
passing  through  the  star;  it  is  reckoned  eastward  from 
0°  to  360°,  or  from  O**  to  24\ 

The  angle  at  the  pole  between  the  hour  circle  of  the 
star  and  the  meridian  of  the  place  is  called  the  hour  angle 
of  the  star. 

The  declination  of  a  star  is  its  distance  from  the  equinoc- 
tial, measured  on  its  hour  circle ;  it  may  be  north  or  south, 
and  is  usually  reckoned  from  0°  to  90°.  It  corresponds  to 
terrestrial  latitude. 

The  poh  "  distance  of  a  star  is  its  distance  from  the  pole, 
and  is  the  complement  of  its  declination.  The  right  ascen- 
sion and  declination  of  celestial  bodies  are  given  in  nautical 
almanacs. 

III.  The  ecliptic  and  the  circle  of  latitude  through  the  vernal 
equinox. 

The  latitude  of  a  star  is  its  angular  distance  from  the 
ecliptic  measured  on  a  circle  of  latitude ;  it  may  be  north 
or  south,  and  is  reckoned  from  0°  to  90°. 

The  longitude  of  a  star  is  the  arc  of  the  ecliptic  inter- 
cepted between  the  vernal  equinox  and  the  circle  of  latitude 
passing  through  the  star. 


] 


)ernal 


the 
north 

inter- 
,titude 


8PBEHICAL  COORDINATES. 


341 


ftt» 


2.^4.  Qraphic  Representation  of  the  Spherical  Coordi- 
nates. —  The  figure  will 
serve  to  illustrate  the  pre- 
ceding definitions.  0  is 
the  earth,  PHP'R  is  the 
meridian,  P  the  north  pole, 
HR  the  horizon,  EQ  the 
equinoctial,  Z  the  zenith. 
Then,  of  a  place  whose 
^nith  is  Z,  QZ  is  the  ter- 

strial  latitude ;  and  since 

QZ  =  PE, 

.*.  PR  =  the  latitude. 

But  PR  is  the  elevation  of  the  pole  above  the  horizon. 

Hence  the  elevation  of  the  pole  above  the  horizon  is  equal  to 
the  latitude. 

Let  V  be  the  vernal  equinox,  and  let  S  be  any  heavenly 
body,  such  as  the  sun  or  a  star ;  then  its  position  is  denoted 
as  follows : 


^©V-^1 


A\ 


VK  =  right  ascension          of  the  t 

)od3 

f  =  a, 

KS  =  declination                  " 

(I 

« 

=  8, 

ZPS  or  QK  =  hour  angle      « 

u 

(( 

=  t, 

PS  =  north  polar  distance  " 

(( 

(( 

=i>» 

HT  =  azimuth                      « 

« 

i( 

=  a, 

TS  =  altitude            '          « 

« 

u 

=  A, 

ZS  ~  zenith  distance           " 

« 

ii 

=  2, 

QZ         H  =  latitude  of  the  observe 

;r 

^4,. 

The  triangle   ZPS   is  called  the  astronomical  triangle; 
ZP  =  90°  —  <^  =  co-latitude  of  the  observer, 

PS  =  90°-8,  SZ  =  90°-A. 


:i:' 


!■, 


342 


SPHF.niCA L  TlilGONOMETIt  Y. 


7 


Let  the  small  circle  MM',  passing  througli  S,  and  parallel 
to  the  equinoctial,  represent  the  apparent  diurnal  iiiotion  of 
the  heavenly  body  S  (the  declination  being  supposed  con- 
stant); then  the  body  S  will  appear  to  rise  at  A  (if  we  sup- 
pose the  Eastern  heiulsphere  is  represented  in  the  diagram). 
It  will  be  at  B  at  6  o'clock  in  the  morning,  at  M  at  noon, 
at  M'  at  midnight,  and  at  <»  it  will  be  east. 

225.  Problems.  —  By  means  of  the  foregoing  definitions 
and  diagram  we  may  solve  several  astronomical  problems  of 
an  elementary  characster  as  follows  : 

(1)  Given  the  IrUtude  of  a  place  and  the  declination  of  a 
star;  to  find  the  time  of  its  rising. 

Let  A  be  the  position  of  the  star  in  the  horizon.  Then 
in  the  triangle  APB,  right  angled  at  It,  we  liave 

cos  RPA  =  -  cos  ZPA  =  tan  KP  cot  AP. 


.*.  cos  t  =  —  tan  <^  tan  S    .    . 
from  which  the  hour  angle  is  found. 


(1) 


Since  the  hourly  rate  at  which  a  heavenly  body  appears 
to  move  from  east  to  west  is  15",  if  the  hour  angle  be 
divided  by  15  the  time  will  be  found.  In  the  case  of  the 
sun,  formula  (1)  gives  the  time  from  sunrise  to  noon,  and 
hence  the  length  of  the  day. 

Ex.  Required  the  apparent  time  of  sunrise  at  a  place 
whose  latitude  is  40°  3G'  23".9,  on  July  4,  1881,  when  the 
sun's  declination  is  22°  52'  1". ' 


<^  =  40°36'23".9, 
8  =  22°  52'    1". 


log  tan  <^  =  9.0.331352 
log  tan  8  =  9.6250362 

log  cos  t  =  9.5581714- 

.-.  t  =111°  11' 44" 

_  71. 24"'  47»,  nearly, 


t 


\ 


(1) 


PROBLEMS  OF  SPIIEHWAL  AnTllONOMY.       o48 

which  taken  from  12'',  the  time  of  apparent  noon,  gives 
4'' 35"'  13",  the  time  of  apparent  sunrise.* 

(2)  Oiven  the  latitude  of  a  place  and  the  declination  of  a 
star;  to  find  its  azimuth  from  the  north  at  rising. 

Let  A  =  the  azimuth  =  All.  Then  in  the  triangle  APR 
we  have 

sin  AP  =  cos  AK  cos  PR, 

or  sin  8   =  cos  A  cos  </>. 

.*.  cos  A=sindsec<^ (2) 

Ex.  Required  the  hour  angle  and  azimuth  of  Arcturus 
when  it  rises  to  an  observer  in  New  York,  lat.  40°  42'  N., 
the  declination  being  19°  57'  N. 

Ans.  7"  12""  46'.3 ;  N.  G3°  15'  11"  E; 

(3)  Oiven  the  latitude  of  the  observer  and  the  hour  angle 
and  declination  of  a  star;  to  find  its  azimuth  and  altitude. 

Here  we  have  given,  in  the  triangle  ZPS,  two  sides  and 
the  included  angle;  that  is,  PZ  =  90°-</),  PS  =  90°  -  8, 
and  ZPS  =  t.  Let  A  =  the  azimuth  from  the  north  =  RT, 
p  =  the  angle  ZSP,  and  z  =  ZS.  Then  by  Delambre's 
Analogies  (Art.  198), 

8in^(p  +  A)  cos  ^ z  =  cos  ^(8  —  <f>)  cos ^ t, 

sm^{p  —  A)  sin  ^«  =  sin  ^(8  —  <^)cos^f, 

cos^(2)-f-  A)  cos -^2  =  sin  1(8  +  ^)sin^^, 

cos^(2>  —  A)  sin  ^z  =  cos|(8  +  ^)  sin^f. 

Hence,  when  <^,  t,  and  8  are  given,  that  is,  the  latitude  of 
the  place,  and  the  hour  angle  and  the  declination  of  a 
heavenly  body.  A,  z,  and  p  can  be  found. 

In  a  similar  manner  may  be  solved  the  converse  problem  : 
Given  the  latitude  of  the  observer  and  the  azimuth  and  altitude 
of  a  star;  to  find  its  hour  angle  and  declination. 

*  In  these  examplea  no  corrections  arc  applied  tdt  refraction,  semi-diameter  of 
the  sun,  change  in  declination  from  noon,  etc. 


44 


SPHERICAL   TRIGONOMETRY. 


^4)   Given  the  right  ascensioiis  and  declinations  of  two  stars; 
to  find  the  distance  between  them. 

Let  P  be  the  pole,  S  and  S'  the  two  stars,    p^a-ar 
Let  «  and  «'  be  the  right  ascensions  of  the 

stars ; 

8  and  8'  their  declinations ;  and 

d  the  required  distance. 

Then  we  have  given,  in  the  triangle  PSS', 
two  sides  and  the  included  angle ;  that  is, 
PS  =  90°-8=i),  PS'  =  90°-8'=i)',  and 
P  =  «  -  a'. 

This  may  be  solved  by  Art.  198,  or  by  the 
second  method  of  Art.  209,  as  follows : 


Draw  SD  perpendicular  to  PS'  produced;   let  PD 
Then 

cosP  =  tanPDcotPS. 


m. 


.*.  tan  m  =  cos  P  tan  p. 
Also  cos  SS'  =  cos  PS  cos  S'D  sec  PD 

.•.    cos  d  =  cosp  cos  S'D  sec  m. 


(Art.  209) 


Ex.  Required  the  distance  between  Sirius  and  Aldebaran, 
the  right  ascensions  being  6"  38""  37'.6  and  4''  27""  25".9,  and 
the  declinations  16°  31'  2"  S.  and  16°  12'  27"  N.,  respectively. 


Here  P  =  2''  11«  11'.7 

=   32°  47' 65" 

j9  =  106°31'   2" 

m  =  109°  25'  55" 

p'=   73°  47' 33" 

S'D=   35°  38' 22" 

j>  =  106°3r   2", 

m  =  109°  25155", 

SS'  =  d=   46°   0'44". 


log  cos  P  =  9.9245789 
log  tan  p  =  0.5279161  - 
log  tan  m  =  0.4524950- 

.-.  m  =  109° 25' 55". 

log  cos  S'D  =  9.9099302 
log  cos  p^- 9. 4537823- 

colog  cos  m  =  0.4779643- 
log  cos  cZ  =  9.8416768 


PROBLEMS  OF  SPHERICAL  ASTRONOMY.       346 


\> 


(5)   Given  the  right  ascension  and  declination  of  a  star;  to 
find  its  latitude  and  longitude. 

Let  V  be  the  vernal  equinox,  S  the  star,  VD,  VL  the 
the  equator  and  the  ecliptic,  SI),  SL 
perpendicular  to  VD,  VL.  Then 
VD  =  right  ascension  =  «,  SD  =  dec- 
lination =  8,  VL  =  longitude  =  X, 
SL  =  latitude  =  I.  Denote  the  ob- 
liquity of  the  ecliptic  DVL  by  w, 
and  the  angle  DVS  by  0. 

From    the    right  triangles    SVD, 
SVL  we  get 

cot  ^  =  sin  rt  cot  8 (1) 

tanX  =  cos  (^  — o))  tanasec^ (2) 

sin  I  =  sin  {6  —  w)  sin  8  cosec  6 (3) 

From  (1),  $  is  determined;  and  from  (2)  and  (3),  X  and 
/  are  determined. 

Ex.  Given  the  right  ascension  of  a  star  5'*  6""  42'.01,  and 
its  declination  45°  51' 20".  1  N. ;  to  find  its  longitude  and 
latitude,  the  obliquity  of  the  ecliptic  being  23°  27'  19". 45. 


«  =  76°40'30".15 
8  =  45°  51'  20".l 
^  =  46°  38' 11  ".8 
23°  27'  19".45 


<i> 


tf-a)  =  23°10'52".35 


A  =  79°  68'   3".44. 


Z=22°51'48".4. 


log  sin  «=    9.9881479 
log  cot  8  =    9.9870277 

log  cot  d  =   9.9751756 

log  cos  (^  -  o>)  =   9.9634401 

log  tan  a  =  10.6255266 

cologcos^=    0.1632816 

log  tan\  =  10.7522483 

log  sin  (^  -  0))  =    9.5950996 

log  sin  8  =    9.8558743 

cologsin^=   0.1384575 

log  sin/  =   9.5894314 


346 


SPIJElilCA  L   TltlOONOMETlt  V. 


EXAMPLES. 

1.  Find  the  apparent  time  of  sunrise  at  a  place  whose 
latitude  is  40"  42',  when  the  sun's  declination  is  17°  49'  N. 

Ana.  4^  CC". 

2.  Given  the  latitude  of  a  place  =  40°  36'  23".9,  the  hour 
angle  of  a  star=4()°40'4".o,  and  its  declination  =  2.'{°  4'  24".3 ; 
to  lind  its  azimuth  and  altitude. 

An8.  Azimuth  =  80°  23'  4".47,  altitude  =  47°  15'  18".3. 

3.  Find  the  altitude  and  azimuth  of  a  star  to  an  observer 
in  latitude  38°  53'  N.,  when  the  hour  angle  of  the  star  is 
3h  15...  20«  w.,  and  the  declination  is  12°  42'  N. 

Ana.  Altitude  =  39°  38'  0";  azimuth  =  S.  72°  28'  14"  W. 

4.  Given  the  latitudes  of  New  York  City  and  Liverpool 
40°  42' 44"  N.  and  53°  25' N.,  respectively,  and  their  longi- 
tudes 74°0'24"W.  and  3°  W.,  respectively;  to  find  the 
shortest  distance  on  the  earth's  surface  between  them  in 
miles,  considering  the  earth  as  a  perfect  sphere  whose 
radius  is  3950  miles. 

NoTB.  —  This  ia  evidently  a  case  of  (4)  where  two  ■ides  and  the  included  angle 
are  given,  to  tind  the  third  aide. 

Ana.  3305  miles. 

6.  The  latitudes  of  Paris  and  Pekin  are  48°  50'  14"  N. 
and  39°  54'  13"  N.,  and  their  difference  of  longitude  is 
114°  7' 30";  find  the  distance  between  them  in  degrees.  ^ 

Ana.  78°  56' 40". 


GEODESY. 

226.  The  Chord al  Triangle.  —  Given  ttvo  aidea  and  the 
included  angle  of  a  apherical  triangle;  to  find  the  correspond- 
ing angle  of  the  chordal  triangle. 


GEODESY. 


847 


14"  N. 
ade  is 
s.  ^ 

;G'40". 


The  chordal  triangle  is  the  triiiuglo  formed 
by  the  chords  of  tlie  sides  of  a  spherical 
triangle. 

Let  ABC  be  a  spherical  triangle,  0  the 
centre  of  the  sphere,  A'BC  the  cohinar 
triangle,  and  M,  N  the  middle  points  of 
the  ares  A'B,  A'C.  Then  the  chord  AB  is 
parallel  to  the  radius  ()M,  since  they  are 
both  perpendicular  to  the  chord  A'P>.  Simi- 
larly, AC  is  parallel  to  ON. 

In  the  spherical  triangle  A'MN,  we  have 

cosMNsscosA'NcosA'M  +  sinA'NsiuA'McosA'  (Art.  191) 

Denote  the  angle  BAC  of  the  chordal  triangle  by  A^ 
Then    arc   MN   or  angle   MON  =  A„    A'N  =  ^(7r-6), 
A'M  =  Ht  -  c),  and  A'  =  A. 

.'.  cos  A,  =  sin^i  sin|^c-f- cos^6  cosl^ccos  A  .     (1) 
with  similar  values  for  cos  B^  and  cos  C^. 

Cor.  1.  If  the  sides  h  and  c  are  small  compared  with 
the  radius  of  the  sphere,  Ai  will  not  differ  much  from  A. 

Let  Ai  =  A  -  ^ ;  then 

cos  A,  =  cos  A  +  ^  sin  A,  nearly. 

But     sin  ^6  sin^c  =  sin^|(6 -f  c)  — sin'^ft  —  c), 

and         cos  ^Z>  cos  ^c  =  (iOii^\{h  +  f,  —  sin'-*  \{h  —  c). 

Substituting  in  (1)  and  reducing,  we  get 

■e  =  tan|Asin2|(i>4-c)-cot^A8in''^(&-c)  .     (2) 

which  is  the  circular  measure  of  the  excess  of  an  angle  of  the 
spherical  triangle  over  the  corresponding  angle  of  the  chordal 
triangle. 

The  value  in  seconds  is  obtained  by  dividing  tlie  circular 
measure  by  the  circular  measure  of  one  se(H)nd,  or,  approxi- 
mately, by  the  sine  of  one  second. 


I  H 


848  SPHERICAL   TRWONOMKTJiY. 

Coi'.  2.  The  anijlex  of  the  chordal  triangle  are,  reftpectively, 
eqnai  to  the  arcs  joining  the  middle  points  of  the  sides  of  the 
colunar  triang^^s. 

22n.  Legendre's  Theorem.  —  If  the  sidps  of  a  spherical 
triangle  be  small  compared  with  the  railius  of  the  sphere,  then 
each  angle  of  the  spherical  triangle  exceeds  by  one-third  of  the 
spherical  excess  the  corresponding  angle  of  the  plane  triangle, 
the  sides  of  which  are  of  the  same  length  as  the  arcs  of  the 
spherical  triangle. 

Let  a,  b,  c  be  the  lengths  of  the  sides  of  the  spherical 
triangle,  and  r  the  radius  of  the  sphere;  then  the  circular 

measures  *  of  the   sides  are  respectively  -,  - ,  —     Hence, 

'/•  r  r 

uegleci.ing  powers  of  -  above  the  fourth, 

r 


C08-  — 

r 

K 

■  cos     COS 

r 

r 

sin 

b  .    c 
-sin- 

r      r 

C08A  = ; (Art.  191) 


Y pT^ pr \^^^-  ^^^) 


be. 


bc\       2»-»  24<«--        A  ^^  J 

\      2bc.r  246a-'.       A         *>»*/ 

=  y  +  c'-q' _ 26«c'  +  2c'a-  4-2a'ft«-a«- 6«- c« 
26c  246cr» 


(1) 


*The  u    1  ?  is  tli>>  "'rrular  meaaiire  of  the  angle  wbicb  tbe  arc  u  aubU  ida  al  tb« 


r  he 

centre  of  tuc  scbere;  and  aiaiilarly  fur  -  nud  .. 

r        r 


GEODESY. 


849 


lively, 
of  the 


kerical 
e,  then 
of  the 
iangle, 
of  the 

herical 
ircular 

Hence, 


t.  191) 


rt.l56) 


.     (1) 


t(  idi  »l  the 


Now  if  A',  B',  C  denote  the  angles  of  the  plane  triangle 
whose  sides  are  a,  b,  c,  respectively,  we  have 

cosA'  =  ^li^'     ....     (Art.%) 


and    .;»»  A  >  -  2&V  4-  2c'a«  •:■  2a«ft'  -a^-b*-c* 


46V 


(Art.  100) 


Therefore  (1)  becomes 


cosA==cosA'-^l^A'     ....     (2) 

Let  A  =.  A'  4-  ^,  where  0  is  a  very  small  quantity  ;  then 

cos  A  =  (!08  X'  —  6  sin  A',  neprly. 

.    /J      6c  sin  A'       A 

•  ^'^~~7n^'^3? (Art.  101) 

where  A  denotes  the  area  of  the  plane  triangle  whose  sides 
are  a,  6,  c. 


We  have  therefore  A  =  A'  + 

A 


3r» 

A 


Similarly      13=:H'  +  A,  C  =  C'4-A. 
.-.  A  +  Ii-fC-A'  ^B'-C'  =  -; 


or 


A-fB  +  C  —  ir=—    =  spherical  excess  (Art.  210) 


.-.  A  — A'=B  — B'=C— ^'=  --^=  J  spherical  excess. 

Cor.  1.  If  the  sides  of  a  spherical  triangle  be  very  small 
compared  xnth  the  radius  of  the  sphere,  the  area  of  the 
spherical  triangle  is  approximoMy  equal  to 

aA4-«^-+A^  +  '^. 

\  24r«      ; 


i 


350 


SPHERICAL   TliiaONOMETliY. 


For,  tau  \  E  =  x/tau  —  tan -—  tan '^  tan '^^^  (Art.  220) 
'        ^         \       2r        2r  2r         2r    ^  ' 


and 


tan  A  =  ^[1+    «n.  tan^=^  =  ^-«ri+ (^^1  etc. 
2r     2rL       12j'2j'  2r         2r  [  lii'''  J 

(Art.  156) 


*         \2r     2r      2/     2r  [       12r'JL  12r  J 


.-.  iE  =  A   /i  +  Z±llZliLt±l^ZliOl±Vii£): 

(Arts.  101  and  156) 

\i\  Vlr'     J      4i\  24:7^      J 

...  E,-  =  Afl+^^Y 

That  is :  ^/le  area  of  the  sphei'kal  triangle  exceeds  the  area 

a^  4.  lyi  4.  c- 
0/  f /ie  j?i(t/ie  triangle  by  — ^>l~? —  i^^'*^  ''Z  ^'^^  Ztt^^er. 


24  7- 


1 


Cor.  2.  If  we  omit  terms  of  the  second  degree  in  -,  we  have 

•  r 

E».2^^. 

Hence,  if  the  sides  of  a  spherical  triangle  be  very  small 
compared  tmth  the  radius  of  the  sphere,  its  area  is  approxi- 
mately  equal  to  the  area  of  the  plane  triangle  having  sides 
of  the  same  length. 

228.  Roy's  Rule.  —  The  are  'j  ^f  a  spherical  triangle  .;n  the 
Earth's  surf((ce  being  known,  to  establish  a  formula  for  com- 
puting t'  8  spherical  excess  in  seconds. 

Let  A  be  the  area  of  the  triangle  in  square  feet,  a  id  n  the 
number  of  seconds  in  the  spherical  excess.     Then  we  have 


ROY'S  RULE. 


351 


A  = 


180°  X  60  X  60 


ni* 


....  (Art.  210) 

180x60x60     206265 ^^^ 

Now,  the  length. of  a  degree  on  tlie  Earth's  surface  is 
found  by  actual  measurement  to  be  365155  feet. 

...   ^=365155.    .,,^180x365155. 

180°  TT 

Substituting  this  value  of  r  in  (1),  and  reducing,  we  get 

log  n  =  log  A -9.3267737 (2) 

This  formula  is  called  General  Roy's  rule,  as  it  was  used 
by  him  in  the  Trigonometric  Survey  of  the  British  Isles. 
He  gave  it  in  the  following  form :  From  the  logarithm  of 
the  area  of  the  triangle^  taken  as  a  j^lane  triangle,  in  .square 
feet,  subtract  the  constant  logarithm  9.3267737;  and  the  re- 
mainder is  the  logarithm  of  the  excess  above  180°,  in  seconds, 
nearly. 

Ex.   If  the  observed  angles  of  a  spherical  triangle  .are 
42°  2' 32",  67°  55' 39",  70°1'48",  and  the  side  opposite  the 
angle  A  is  27404.2  feet,  required  the  number  of  seconds  in 
the  sum  of  the  errors  made  in  observing  the  three  angles. 
Here  the  ajyparent  spherical  excess  is 

A  +  B  +  C  -  180°  =  -  1". 
The  area  of  the  triangle  is  calculated  from  the  expression 
a^  sin  B  sin  0 


2  sin  A 


(Art.  101) 


and  by  Roy's  Rule  the  computed  spherical  excess  is  found 
to  be  .23". 

Now  since  the  computed  spherical  excess  may  be  suppose  _i 
to  be  the  real  spherical  excess,  the  sum  of  the  obser/ed 
angles  ought  to  have  been  180°  -f-  .23". 

Hence  it  appears  thnt  the  sum  of  the  errors  of  the  obser- 
vations is  .23"  — (— 1")  =  1".23,  which  the  observer  must 


m 


!     I 


«^AM*MUM»ie 


352 


SPHERICAL   TRIGONOMETRY. 


add  to  the  thiee  observed,  angles,  in  such  proportions  as  his 
judgment  may  direct.  One  way  is  to  increase  each  of  the 
observed  angles  by  one-third  of  1".23,  and  take  the  angles 
thus  corrected  for  the  true*  angles. 


•JV 


229.  Reduction  of  an  Ang^le  to  the  Horizon.  —  Given  the 
angles  of  elevation  or  dejrreKsion  ofttvo  objects,  which  are  at  a 
small  angular  distance  from  the  horizon,  and  the  angle  which 
the  objects  subtend,  to  find  the  horizontal  angle  between  them. 

Let  a,  b  be  the  two  objects,  the  angular  distance  between 
which  is  ireasured  by  an  observer  at 
0 ;  let  ()Z  be  the  direction  at  right 
angles  to  the  observer's  horizon.  De- 
scribe a  sphere  round  0  as  a  centre, 
and  let  vertical  planes  through  Oa, 
06,  meet  the  horizon  at  OA,  OB,  re- 
spectively ;  then  the  horizontal  angle 
AOB,  or  AB,  is  required. 

Let    ab  =  0,    AI>  =  ^  -f  a;.    Art  =  h, 
B6  =  k.     Then  in  the  triangle  aZb  we  have 


cos  AB  =  cos  aZb  = 


cos  ab  —  cos  aZ  cos  6Z 


or 


,a,      V       cos^ 
COS  {$  +  a;)  =  - 


sin  aZ  sin  6Z 
s'lnh  sink 


cos  h cos  k 


This  gives  the  exact  value  of  AB ;  by  approximation  we 
obtain,  where  x  is  essentially  small, 

y,  .    /I         cos  6  —  hk 

.*.  a;  sin  ^  =  hk  —  |  {h^  +  fc*)  cos  $,  nearly. 

e 


2  hk  -  {h'  +  k')  ( cos^  ;^  -  sin2 


x  = 


2sin0 
=  \i{h  4  A;)' tan  ^ e -  {h  -  ky  cot^  ^]. 


SMALL   VAUIATlOys  IN  PA  UTS  OF  TRIANGLES.     353 


1 


EXAMPLES. 

1.  Prove  that  the  angles  subtended  by  the  sides  of  a 
spherio'.l  triangle  at  the  pole  of  us  circunicircle  are  respec- 
tively double  the  corresponding  angles  of  its  chordal  tri- 
angle. 

2.  If  Ai,  B|,  Ci ;  Aj},  B^  ^2  >  A.,,  B3,  C;j ;  be  the  angles  of 
the  chordal  triangles  of  the  colunars,  prove  that  , 

co8Aj=co8ja8inS,  co8B,  =  8in  J/>8iu(iS— C),  co8C,=8liUc8in(S— B), 

C08A,=8iiUa8m(S  — C),co8B,=co8.j68iaS,  co8C,=8in  Jc8in(S- A), 

co8A3=siniasin(S— B),  co8B3=8inA^8in(S— A),co8C3=:co8ic8inS. 

3.  Prove  Legendre's  Theorem  from  either  of  the  formulae 
for  sin  ^  A,  cos  ^  A,  tan  ^  A,  respectively,  in  terras  of  the 
sides. 

4.  If  C  =  A  -f  B,  prove  cos  C  =  —  tan  ^  a  tan  ^  6. 

230.  Small  Variations  in  the  Farts  of  a  Spherical  Tri- 
angle. 

It  is  sometimes  important  in  Geodesy  and  Astronomy  to 
determine  the  error  introduced  into  one  of  the  computed 
parts  of  a  triangle  from  any  small  error  in  the  given  parts. 

If  two  parts  of  a  spherical  triangle  remain  constant,  to  deter- 
mine the  relation  between  the  small  variations  of  any  other  two 
parts. 

Suppose  C  and  c  to  remain  constant. 

(1)  Required  the  relation  between  the  small  variations 
of  a  side  and  the  opposite  angle  (a,  A). 

Take  the  equation 

sin  A  sine  =  sin  C  sin  a (1) 

We  suppose  a  and  A  to  receive  very  small  increments  da 
and  dA;  then  we  require  the  ratio  of  da  and  dA  when  bo'h 
are  extremely  small.     Thus 

sin  (A  4-  dA)  sin  c  =  sin  C  sin  (a  +  da), 


IM^4 


864  SPHERICAL   TRIGONOMETRY. 

or  (sin  A  cos  dA  +  cos  A  sin  t/A)  sin  c 

=  sin  C  (sin  a  cos  iki  -f  cos  a  sin  da) (2) 

Because  the  arcs  dk.  and  da  are  extremely  small,  tlieir 
sines  are  equal  to  the  arcs  themselves  and  their  cosines 
ccjual  1 :  therefore  (2)  may  be  written 

sin  A  sin  c  -f  cos  A  sin  cdX  =  sin  C  sin  a  4-  sin  C  cos  ada     (3) 

Subtracting  (1)  from  (3),  we  have 

cos  A  sin  cdA  =  sin  C  cos  ada. 

da  _  cos  A  sin  c  _  tan  a 
dA.     sin  C  cos  a     tan  A 

(2)  Required  the  relation  between  the  small  variations 
of  the  other  sides  («,  b).     We  have 

cos  c  =  c»s  a  cos  6  + sin  a  sin  6  cos  C (1) 

.*.  co8c  =  cos(a+da)cos(6+(/6)  +sin(a+<?a)sin(64-rf6)cosC, 

or  =  (cos  a  —  sin  ada)  (cos  h  —  sin  hdh) 

+  (sin  a  +  cos  ada)  (sin  6  +  cos  hdh)  cos  C  .     (2) 

Subtracting  (2)  from  (1)  and  neglecting  the  product 
da  dbf  we  have 

0  =  (sin  a  cos  b  —  cos  a  sin  b  cos  C)  da 

-f  (cos  a  sin  b  —  sin  a  cos  b  cos  C)  db, 

0  _(cot&sina— cosacosCK       (cotasin^>— cosfecosC) ,. 
sin  a  sin  b 

n     cot  B  sin  C  da  ,  cotAsinCdft  ,»   ,    ^nox 

0  = : 1 ; ....      (Art.  I'Jo) 

Sin  a  sm  b 

da__  cos  A 
db         cos  B 

(',<)  Required  the  relation  between  the  small  variations 
of  the  other  angles  (A,  B). 


SMALL  VARIATIONS  IN  PARTS  OF  TRIANGLES.      365 


.     (2) 

I,  their 
cosines 

a     (3) 


nations 

.     (1) 
lb)cos C, 

D.     (2) 
product 

!0s  C)  dh, 
A.rt.  193) 


variations 


By  means  of  the  polar  triangle,  we  may  deduce  from  the 
result  just  found,  that 


dB 


cos^ 
COS  b 


(4)  Required  the  relation  between  the  small  variations 
of  a  side  ami  the  adjacent  angle  {b,  A).     We  have 

cot  c  sin  6  =  cot  C  sin  A  +  cos  6  cos  A  .     .     .     (Art.  103) 

Giving  to  b  and  A  very  small  increments,  and  subtracting, 
as  before,  we  get 

cote  cos  6r7/>  =  cot  C  cos  Ad  A  —  sinftcos  Art6  —  cosftsin  AdA. 
(cote  cos  b  4-  sin  b  cos  \)db==  (cot  C  cos  A  —  cos  b  sin  A)dX. 


cos  a  ,,          cos  B  , 
.'.  — : —  db  = da 

sine  sinC 


db 

cos  B  sin  6 

dA 

cos  a  sin  B 

EXAMPLES 

(Arts.  191  and  192) 
sin  6cotB 


cos  a 


1.  If  A  and  c  are  constant,  prove  the  following  relations 
between  the  small  variations  of  any  two  parts  of  the  other 
elements : 


da 
dC 

dh 
dC 


tana 
tanC 

tana 


sma 


db 


dB     sin  C 


dC 
dB 


cos  a. 


sinC  ' 

2.    If  B  and  C  remain  constant,  prove  the  following: 
db      tan  b  dA 


do       tan  c ' 

dA       •     A  i.      I. 
— —  =  sin  A  tan  b ; 

do 


da 

da 
do 


=  sin  B  sin  C. 


sma 
sin  c  cos  b 


I 


356 


SPHERICAL   TRIGONOMETRY. 


POLYEDRONS. 

231.  To  find  the  Inclination  of  Two  Adjacent  Faces  of  a 
Eegnlar  Folyedron. 

Let  C  and  D  be  the  centres  of  the  cir- 
cles inscribed  in  the  two  adjacent  faces 
whose  common  edge  is  AB ;  bisect  AB 
in  E,  and  join  CE  and  DE ;  CE  and  DE 
will  be  perpendicular  to  A]i.  .•.  Z  CED 
is  the  inclination  of  the  two  adjacent 
faces,  which  denote  by  I. 

In  the  plane  CED  draw  CO  and  DO  at 
right  angles  to  CE  and  DE,  respectively, 
and  meeting  in  0.  Join  OA,  OE,  OB,  and  from  0  as  centre 
describe  a  sphere,  cutting  OA,  OC,  OE  at  a,  c,  e,  respec- 
tively ;  then  ace  is  a  spherical  triangle.  Since  AB  is  per- 
pendicular to  CE  and  I'^E,  it  is  perpendicular  to  the  j)lane 
CED;  therefore  the  plane  AOB,  in  which  AB  lies,  is  per- 
pendicular to  the  plane  CED.     .•.  Z  aec  is  a  right  angle. 

Let  m  be  the  number  Ox  sides  in  each  face,  and  n  the 
number  of  plane  angles  in  each  solid  angle.     Then 

Zace=ZACE  =  -?^  =  -51. 


2m 


m 


and 


/.cae  =  ^Z  of  the  planes  OAC  and  OAD. 


.*.  Z  cae  = 


2Tr 
2n 


IT 

n 


In  the  right  triangle  cae  we  have 
cos  cae  =  cos  ce  sin  ace. 

But  ^os  ce  =  cos  COE  =  cos  ( ^  —  - 

IT        •    T    •    ir 
.•.  cos- =  sm-sin— 

n  2      m 

.'.  8in-  =  co8-cosec— • 
2  n  m 


=  sin -• 


VOLUME  OF  A   PAIiAlLELOPlPED. 


357 


Cor.  1.    Ifrbe  the  radius  of  the  inscribed  sphere,  and  a  be 
a  side  of  one  of  the  faces,  then 


r  = -cot— tan  — 
2       m       2 


For, 


r  =  OC=CEtanCEO=AEcotACEtanOEO 


=  „  cot     tan  — 


2       m       2 
Cor.  2.   IfRbe  the  circumradius  of  the  polyedron,  then 

R  =  -  tan-tan   • 
2       n       2 

For,  r  =  C)  A  cos  aoc  =  K  cot  eca  cot  eac  =  R  cot  "^  cot  -  • 

»u       n 

.'.  R  s=  -  tan- tan-- 


2 


2 


Cor.  3.    The  surface  of  a  regtdar  polyedron,  F  being  the 

number  of  faces,   —  '^^i^i-i.  cot  — • 

4  m 

For,  the  area  of  one  face  =  "  w  cot—-     .-.  etc. 

4  m 

Cor.  4.    The  volume  of  a  regular  polyedron 

ma'rF     .  rr 
^  cot— • 


12 


m 


For,  the  volume  of  the  pyramid  which  has  one  face  of 
the  polyedron  for  base  and  O  for  vertex 

~o*~T~^ot  — -     .'.  etc. 
o     4         m 


232.  Volume  of  a  Parallelopiped.  —  To  find  the  volume  of 
a  parallelopiped  in  terms  of  its  edges  and  their  incHnations 
to  one  another. 


358 


SPHERICAL   THWONOMETRY. 

Q 


Let  the  edges  be  OA  =■  a, 
OB  =6,  OC  =  c,  and  let  the 
inclinations  l)e  BOC  =  «,  COA 
=  /3,  AOB  =  y.  Draw  CH 
perpendicular  to  the  face 
AOBE.  Describe  a  Sphere 
round  O  as  centre,  meeting 
OA,  OB,  00,  OE,  in  a,  b,  c,  e, 
respectively. 

The  volume  of  the  parallelo- 
piped  is  equal  to  the  area  of  tlie  base  OAEB  multiplied  by 
the  altitude  CH ;  that  is, 

volume  =  ab  sin  y  •  CH  =  abc  sin  y  sin  ce 

where  ce  is  the  perpendicular  arc  from  c  on  ab. 

.'.  volume  =  a6c  sin  y  sin  etc  sin  6ac     .    .     .     (Art.  186) 

2n 


=  a6c  sin  y  sin  ^ 


sin  /3  sin  y 


(Art.  195) 


=  abc  Vl  —  cos' «  —  cos"  /8  —  (los"  y  -H  2  008  a  cos  ft  cos  y. 
Cor.  1.    The  surface  of  a  paralklopiped 
—  2  {be  sin  «  4-  ca  sin  /3  +  at  sin  y) . 

Cor.  2.    The  volume  of  a  tetraedron 

= ^  abc  Vl — cos'  a — cos*  0— cos'  y + 2  cos  «  cos  ft  cos  y. 

For,  a  tetraedron  is  one-sixth  of  a  parallelopiped  which 
has  the  same  altitude  and  its  base  double  that  of  the 
tetraedron. 

233.  Diagonal  of  a  Parallelopiped. —  To  find  the  diagonal 
of  a  parallelopiped  in  terms  of  its  edges,  and  their  mutual 
inclinaiions. 

Let  OD  (figure  of  Art.  232)  be  a  parallelopiped,  whose 
edges  OA  =  a,  OB  =  b,  OC  =  c,  and  their  inclinations  BOC 
=  «,  COA  =  ^,  AOB  =  y ;  let  OD  be  the  diagonal  required. 


TABLE  OF  FORMULA. 


859 


and  OE  the  diagonal  of  tho  face  OAB.     Then  the  triangle 
OKD  gives 

OD* «  UE'  -f  ED*  4-  2  OE .  ED  cos  COE 

«  rt»  -f  &»  -H  2  o6  cos  y  4-  e'  -\-2c'  OE  cos  COE       (1) 

Now,  it  is  clear  that  OE  cos  COE  is  the  projection  of  OE 
on  the  line  OC,  and  therefore  it  mnst  l)e  equal  to  the  sum 
of  the  projections  of  OB  and  BE  (or  of  OB  and  OA),  on 
the  same  line.* 

.-.  OE  cos  COE  =  b  cos  «  -|-  a  cos  /8, 

which  in  (1)  gives 

OD'  =  a'  +  Ir  4-  c*  4-  26c  cos«  4-  2ca  cos  ft  4-  2a6  cos  y  .     (2) 

234.  Table  of  FormulsB  in  Spherical  Trigonometry.  —  For 

the   convenience  of  the   student,   uiany  of  the  preceding 
formulae  are  summed  up  in  the  following  table : 

1.  cose  —cos  a  cos  6 (Art.  185) 

2.  sin  6  =  sin  B  sine. 

3.  sin  a  =  sin  A  sine. 

4.  cosC  = -cos  AcosB (Art.  189) 

5.  sin  B  =  sin  b  sin  C. 

6.  sin  A  =  sin  a  sin  C. 

-     sin  a      ^\n  b       sin  c 


(Art.  190) 
(Art.  191) 


sin  A     [sin  B      sin  C 

8.  cos  a  =eos  6  cose  4- sin  6  sine  cos  A  . 

9.  cost  =  cos  c  cos  a  4- sine  sin  a  cos  B. 

10.  cos  c  =  cos  a  cos  6  4-  sin  a  sin  b  cos  C. 

11.  cos  A  =  —cos  B  cos  0  4-  sin  B  sin  C  cos  a       (Art.  192) 

12.  cos  B  =  —  cos  C  cos  A  4-  sin  C  sin  A  cos  b. 

*  From  the  nature  of  projecUona  (Plane  and  Solid  Qeoiii.,  Art.  3'26). 


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360  SPHERICAL   TRIGONOMETRY. 

13.  cos  C  =  —  cos  A  cos  B  4-  sin  A  sin  B  cos  c. 

14.  cot  a  sin  &  =  cot  A  sin  C  +  cos  C  cos  b  .     .     (Art.  193) 
16.  cot  a  sin  c  =  cot  A  sin  B  +  cos  B  cos  c. 

16.  cot  b  sin  a  =cot  B  sin  C  +  cos  C  cos  a. 

17.  cot  b  sin  c  =  cot  B  sin  A  +  cos  A  cos  c. 

18.  cot  c  sin  a  =  cot  C  sin  B  +  cos  B  cos  a. 

19.  cot  c  sin  6  =  cot  C  sin  A  -f  cos  A  cos  b. 

20.  si n  a  cos  B  =  cos  &  sin  c  —  sin  b  cos  c  cos  A   (Art.  194) 

21.  sin  a  cos  C  =  sin  b  cos  c  —  cos  6  sin  c  cos  A. 

22.  sin  b  cos  A  =  cos  a  sin  c  —  sin  a  cos  c  cos  B. 

23.  sin  b  cos  C  =  sin  a  cos  c  —  cos  a  sin  c  cos  B. 

24.  sin  c  cos  A  =  cos  a  sin  6  —  sin  a  cos  b  cos  C. 

25.  sin  c  cos  B  =  sin  a  cos  6  —  cos  a  sin  b  cos  C. 

26.  sin^-A=./^i5llESMfEiI     .     .     .     (Art.  195) 

\  sin  6  sin  c 


nn  1  A         /sin  s  sin  (s  —  a) 

27.    COSiA=\/ — -r- — ^^ '-' 

\       sin  &  sin  c 


28.    taniA=J«Hilz:i^l^(l=l^ 
\      sins  sin  (s  —  a\ 


sin  5  sin  {s  —  a) 

OQ    «i    A     —  ^  Vsin  s  sin  (g  —  a)  sin  (s  — 6)  sin  (s  —  c) 

sin  b  sin  c 
^      2n 

sin  &  sin  c 
where  n  =  Vsin  s  sin  (s  —  a)  sin  (s  —  6)  sin  (s  —  c). 


30.   sinAa=J-^«AS_««^l(SriAJ      ^     ^     ^     (Art.  196) 
\  sin  B  sin  C 


31.  cosla=>l-(S-rMc5?.(S:rC). 
^         \  sin  B  sin  C 


/ 


il 


rt.  193) 


.rt.  194) 


\.it.  195) 


s  —  c) 


Art.  196) 


TABLE  OF  FORMULA. 


361 


32.   tan 


i«=V- 


COS  8  cos  (S  —  A) 


cos  (S  -  B)  cos  (S  -  C) 


33    sin  a     =  2V-co38cos(8-A)  cos(S-^)cos(S-0) 

sin  B  sin  C 

34.  tani(A  +  B)^52iM^L=l|lcot|C    .     .     (Art.  197) 

•   cos  ^  (a +  6)  ^ 

35.  tanKA-B)  =  ^^"^^"~-^cot^C. 

^^  ^      sin^(a  +  6)        ^ 

36.  tan  A  (a +  6)   = -— ^^^  ~^Uan  Ac. 

^^  ^       cos^A  +  B)       ^ 

37.  tan  Ha -6)    ^  gl!li(A  -  B)  ^^^  . 

38.  sinHA+B)oos^c  =  cos|(a-6)cos|C  (Art.  198) 

39.  sin  ^A  —  B)  sin  ^c=  sin  ^{a  —  b)  cos|C. 

40.  cos  ^A  4-  B)  cos  ^c  =  cos  ^(a  +  6)  sin  ^C. 

41.  cos^A  —  B)  sin  |c  =  sin  |(a  +  6)  sin  JC. 


42.    tan  r 


Vsin  (s  —  a)^in  (s  —  ft)  sin  (g  —  c) 
sins 


?i 


sins 


(Art.  215) 


43. 


tanR  =  -—  [sin(s— a)+sin(.s— 6)+sin(s— c)— sins] 

(Art.  217) 


E 


44.    K  =  areaof  A  =  -^7rr2 (Art.  219) 

loO 


45.   sin^E 


n 


2  COB  ^  a  cos  ^  &  cos  ^  c 


.     .     .     (Art.  220) 


46.    tan  |  E  =  Vtanistan^(s— a)  tan^  (••*  —  &)  tan^s— c). 


362  SPHERICAL   TRIGONOMETRY. 

EXAMPLES. 

1.  Find  the  time  of  sunrise  at  a  place  whose  latitude  is 
42°  33'  N.,  when  the  sun's  declination  is  13°  28'  N. 

Alls.  5'^9"'13'. 

2.  Find  the  time  of  sunset  at  Cincinnati,  lat.  39°  6'  N., 
when  the  sun's  declination  is  15°  56'  S.*  Ans.  5"*  6"". 

3.  Find  the  time  of  sunrise  at  lat.  40°  43'  48"  N.,  in  the 
longest  day  in  the  year,  the  sun's  greatest  declination  being 
23°  27'  K  Ans.  4»>  32™  16'.4. 

4.  Find  the  time  of  sunrise  at  Boston,  lat.  42°21'N,, 
when  the  sun's  declination  is  8°  47'  S.  Ans.  &"  14°'. 

5.  Find  the  length  of  the  longest  day  at  lat.  42°  16'  48".3 
N.,  the  sun's  greatest  declination  being  23°  27'  N. 

Ans.  IS'' 5"  50*. 

6.  Find  the  length  of  the  shortest  day  at  New  Bruns- 
wick, N.J.,  lat.  40°  29'  52".7  N.,  the  sun's  greatest  declina- 
tion being  23°  27'  S. 

7.  Find  the  hour  angle  and  azimuth  of  Antares,  declina- 
tion 26°  6'  8.,  when  it  sets  to  an  observer  at  Philadelphia, 
lat.  39°  57'  N.  Ans.  4"  23"'  5'.7  ;  S.  54°  58'  44"  W. 

8.  Find  the  hour  angle  and  azimuth  of  the  Nebula  of 
Andromeda,  declination  40°  35'  N.,  when  it  rises  to  an  ob- 
server at  New  Brunswick,  N.  J.,  lat.  40°  29'  52".7  N. 

9.  Find  the  azimuth  and  altitude  of  Regulus,  declination 
16°  13'  N.,  to  an  observer  at  New  York,  lat.  40°  42'  N.,  when 
the  star  is  three  hours  east  of  the  meridian. 

Ans.  Azimuth  =  S.  71°  12'  30"  E. ;  Altitude  =  44°  10'  33". 

10.  Find  the  azimuth  and  altitude  of  Fomalhaut,  dec- 
lination 30°  25'  S.,  to  an  observer  in  lat.  42°  22'  N.,  when  the 
star  is  2''  5""  36'  east  of  the  meridian. 

Ans.  Azimuth  =  S.  27°  18'  40"  E. ;  Altitude  =  11°  41'  37". 


EXAMPLES. 


363 


13' 


*  ^  11.    Find  the  azimuth  and  altitude  of  a  star  to  an  observer 

in   lat.   39°  57'  N.,    when   the   hour  angle   of   the   star  is 
5h  ijm  40S  east,  and  the  declination  is  62°  33'  N. 

Ans.  Azimuth  =  N.  35°  54'  E. ;  Altitude  =  30°  24'. 

12.  Find  the  hour  angle  {t)  and  declination  (S)  of  a  star 
to  an  observer  in  lat.  40°  36'  23",9  N.,  when  the  azimuth  of 
the  star  is  80°  23'  4".47,  and  the  altitude  is  47°  15'  18".3. 

Ans.  « =  46°  40' 4".53 ;  8  =  23°  4' 24".33. 

13.  Find  the  distance  between  Regulus  and  Antares,  the 
right  ascensions  being  10"  0'"  29'.11  and  16"  20'»  20».35,  and 
the  polar  distances  77°  18'  41".4  and  116°  5'  55".5. 

Ans.  99°55'44".9. 

14.  Find  the  distance  between  the  sun  and  moon  when 
the  right  ascensions  are  12"  39'"  3».22  and  6"  55"*  32'. 73,  and 
the  declinations  9°  23'  16".7  S.  and  22°  50'  21".9  N. 

Ans.  89°52'55".5. 

15.  Find  the  shortest  distance  on  the  earth's  surface,  in 
miles,  from  New  York,  lat.  40°  42'  44"  N.,  long.  74°  0'  24"  W., 
to  San  Francisco,  lat.  37°  48'  N.,  long.  122°  23'  W. 

Ans.  2562  miles. 

16.  Find  the  shortest  distance  on  the  earth's  surface 
from  San  Francisco,  lat.  37°48'N.,  long.  122°  23' W.,  to 
Port  Jackson,  lat.  33°  51'  S.,  long.  151°  19'  E. 

Ans. ;  6444  nautical  miles. 

17.  Given  the  right  ascension  of  a  star  10"  l"  9'.34,  and 
its  declination  12°  37'  36".8  N. ;  to  find  its  latitude  and  longi- 
tude, the  obliquity  of  the  ecliptic  being  23°  27'  19".45. 

Ans.  Latitude  =s  ;  Longitude  =  , 

18.  Given  the  obliquity  of  thei  ecliptic  w,  and  the  sun's 
longitude  A;  to  find  his  right  ascension  a  and  decliiiation  8. 

Ans.  tan  a  =  cos  <o  tan  A, ;  sin  8  =  sin  <a  sin  A. 


364  SPHERICAL   TRIGONOMETRY. 

19.  Given  the  obliquity  of  the  ecliptic  23°  27'  18".5,  and 
the  sun's  longitude  59°  40'  1".6 ;  to  find  his  right  ascension 
(a),  and  declination  (8). 

Ans.  a  =  Sh  49"»  52'.62 ;  8  =  20°  5'  33".9  N. 

20.  Given  the  sun's  declination  1G°  0'  5C".4  N.,  and  the 
obliquity  of  the  ecliptic  23°27'18".2;  to  find  his  right 
ascension  (a),  and  longitude  (X). 

Ans.  a  =  9n4'"19^2;  X  =  136°  7' 6".5. 

21.  Given  the  sun's  right  ascension  14*^  8""  19'.06,  and  the 
obliquity  of  the  ecliptic  23°  27'  17".8 ;  to  find  his  longitude 
(\),  and  declination  (8). 

Ans.  X  =  214°20'34".7;  8  =  12°  58' 34".4  S. 

22.  Given  the  sun's  longitude  280°  23' 52".3,  and  his 
declination  23°  2'  52".2  S. ;  to  find  his  right  ascension  (a). 

Ans,  «  =  18M5"' 14'.7. 

23.  In  latitude  45°  N.,  prove  that  the  shadow  at  noon  of 
a  vertical  object  is  three  times  as  long  when  the  sun's  dec- 
lination is  15°  S.  as  when  it  is  15°  N. 

24.  Given  the  azimuth  of  the  sun  at  setting,  and  also  at 
6  o'clock ;  find  the  sun's  declination,  and  the  latitude. 

25.  If  the  sun's  declination  be  15°  N.,  and  length  of  day 
four  hours,  prove  tan  <f>  —  sin  60°  tan  75°. 

26.  Given  the  sun's  declination  and  the  latitude ;  show 
how  to  find  the  time  when  he  is  due  east. 

27.  If  the  sun  rise  northeast  in  latitude  <^,  prove  that  cot 
hour  angle  at  sunrise  =  —  sin  <f>. 

28.  Given  the  latitudes  and  longitudes  of  two  places ; 
find  the  sun's  declination  when  he  is  on  the  horizon  of  both 
at  the  same  instant. 

29.  Given  the  sun's  declination  8,  his  altitude  h  at 
6  o'clock,  and  his  altitude  h'  when  due  east;  prove 
sin'^  8  =  sin  h  sin  h'. 


EXAMPLES. 


365 


18".5,  and 
ascension 

5'  33".9  N. 

'.,  and  the 
his   right 

56°  T  6".5. 

)6,  and  the 
longitude 

58'  34".4  S. 

>,   and  his 
sion  (a). 
^  45"'  14».7. 

a.t  noon  of 
sun's  dec- 

md  also  at 
tude. 

^th  of  day 
;ude;  show 
(ve  that  cot 


wo  places ; 
zon  of  both 


itude   h  at 
ast ;    prove 


30.  Given  the  declination  of  a  star  30°;  find  at  what 
latitude  its  azimuth  is  45°  at  the  time  of  rising. 

31.  Given  the  sun's  declination  8,  and  the  latitude  of  the 
place  <^ ;  find  his  altitude  when  due  east. 

32.  Given  the  declinations  of  two  stars,  and  the  differ- 
ence of  their  altitudes  when  they  are  on  the  prime  vertical ; 
find  the  latitude  of  the  place. 

33.  If  the  difference  between  the  lengths  of  the  longest 
and  shortest  day  at  a  given  place  be  six  hours,  find  the 
latitude. 

34.  If  the  radius  of  the  earth  be  4000  miles,  what  is  the 
area  of  a  spherical  triangle  whose  spherical  excess  is  1°  ? 

35.  If  A",  B",  C"  be  the  chordal  angles  of  the  polar 
triangle  of  ABC,  prove 

cos  A"  =  sin  I A  cos  [s  —  a),  etc. 

36.  If  the  area  of  a  spherical  triangle  be  one-fourth  the 
area  of  the  sphere,  show  that  the  bisector  of  a  side  is  the 
supplement  of  half  that  side. 

37.  If  the  area  of  a  spherical  triangle  be  one-fourth  the 
area  of  the  sphere,  show  that  the  arcs  joining  the  middle 
points  of  its  sides  are  quadrants. 

38.  Given  the  base  and  area ;  show  that  the  arc  joining 
the  middle  points  of  the  sides  is  constant ;  and  if  it  is  a 
quadrant,  then  the  area  of  the  triangle  is  irr'K 

39.  Two  circles  of  angular  radii,  a  and  (S,  intersect 
orthogonally  on  a  sphere  of  radius  r;  find  in  any  manner 
the  area  common  to  the  two. 

40.  If  E  be  the  spherical  excess  of  a  triangle,  prove  that 
|E  =  tan|a  tan^&  sinC  -  |(tania  tan^&)2siu2C  +  etc. 

41.  Show  that  the  sum  of  the  three  arcs  joining  the 
middle  points  of  the  sides  of  the  colunars  is  equal  to  two 


366  SPHERICAL   TRIGONOMETRY. 

right  angles,  the  sides  of  the  original  triangle  being  regarded 
as  the  bases  of  the  colunars. 

42.  Prove  that 

cos^^a  sin^S  +  sin»^  ft  sin^CS  -  C)  +  sin^J-c  sin^CS  -  B) 
+  2cos  |a  sin  |&  sin|c  sin  S  sin(S  —  B)  sin(S  —  C)  =  1. 

43.  Having  given  the  base  and  the  arc  joining  the  middle 
points  of  the  colunar  on  the  base,  the  circumcircle  is  fixed. 

44.  Prove  sin^6  sin^c  sin(S  -  A)  +  cos^6  cos^csinS 

=  cos^a. 

45.  If  A  4-  B  +  C  =  2  TT,  prove  that 

and  cos  C  =  —  cot  |  a  cot  ^  b. 

46.  Solve  the  equations, 

sin  &  cos  c  sin  Z  +  sin  c  cos  6  sin  Y  =  sin  a, 
sin  c  cos  a  sin  X  +  sin  a  cos  c  sin  Z  =  sin  &, 
sin  a  cos  b  sin  Y  +  sin  &  cos  a  sin  X  =  sin  c, 
for  sin  X,  sin  Y,  and  sin  Z. 

47.  If  6  and  c  are  constant,  prove  the  following  relations 
between  the  small  variations  of  any  two  parts  of  the  other 
elements  of  the  spherical  triangle  ABC  : 

^^tanB.  i^  =  -sin.xtanC; 

dC      tanC  dB 

da        ■    1,    •  dA  sin  A      . 

—  =  sin  B  sm  c ;  "tft  " ^~~5 n ' 

dA  dB         smBcosC 


da 


=  —  sin  a  tan  B 


dA  sin  A 


.—   Olll  tt    vein  A^    5  -  ^^    -f-fc        •         /-t 

dC  dG         cos B  sin 0 

48.   If  A  and  c  remain  constant,  prove  the  following : 

da^sina  ^  =  cos  C. 

dB     tanC  db 


regarded 


S-B) 
C)=l. 

;he  middle 
e  is  fixed. 

^csinS 
=  cos^a. 


a  6, 

QC, 


g  relations 
f  the  other 


lowing : 


EXAMPLES. 


867 


49.    If  B  and  C  remain  constant,  prove  the  following 


db 


=  sin  A  tan  c ; 


da 
db 


sin« 


sin  b  cos  c 


50.   If  A  and  a  remain  constant,  prove  the  following 

^  =  ^"'^"  ^  .  ^  _  tan  c  , 

dB      tan  B  '  dC~  tan  C  ' 

rfft  _  _  cosB  .  db___      sin  b 

dc 


cos  C 


dC 


tan  B  cos  c 


51.  Two  equal  small  circles  are  drawn  touching  each 
other;  show  that  the  angle  between  their  planes  is  twice 
the  complement  of  their  spherical  radius. 

52.  On  a  sphere  whose  radius  is  r  a  small  circle  of  spher- 
ical radius  6  is  described,  and  a  great  circle  is  described 
having  its  pole  on  the  small  circle ;  show  that  the  length 


of  their  common  chord  is 


2r 
sinO 


V-cos2^. 


53.  Given  the  base  c  of  a  triangle,  and  that 

tan  ^a  tan  ^  6  =  tan^^  B, 
B  being  the  bisector  of  the  base,  find  a  —  6  in  terms  of  c. 

54.  If  C  =  A  -I-  B,  show  that 

1  —  cos  a  —  cos  b  +  cos  c  =  0, 

55.  If  A  denote  one  of  the  angles  of  an  equilateral  tri- 
angle, and  A'  an  angle  or  its  polar  triangle,  show  that 

cos  A  cos  A'  r=  cos  A  +  cos  A'. 

56.  Show  that 

cos  a  cos  B  —  cos  b  cos  A      cos  C  -f  cos  c 


sin  a  —  sin  b 


sine 


57.   Prove 


cos  A  =  ^os  g  sin  6  —  sin  a  cos  b  cos  0 

sine 


and      cos  A  +  cos  B  =  2^(^ +  &)  sin^O. 

sine 


368  SPHERICAL   TltlGONOMETRY. 

58.  Prove  Legeiidre's  Theorem  by  means  of  the  relations 

sin  A  _  sinB  _  sinC  . 
sin  a      sin  6       sin  c 

59.  Two  places  are  situated  on  the  same  parallel  of  lati- 
tude <f> ;  iind  the  difference  of  the  distances  sailed  over  by 
two  ships  passing  between  them,  one  keeping  to  the  great 
circle  course,  the  other  to  the  parallel ;  the  difference  of 
longitude  of  the  places  being  2X. 

Ans.  2r[A.cos  <^  — sin~'(co8^sin  X)]. 

60.  If  the  sides  of  a  triangle  be  each  60°,  show  that  the 
circles  described,  each  having  a  vertex  for  jwle,  and  passing 
through  the  middle  points  of  the  sides  which  meet  at  it, 
have  the  sides  of  the  supplemental  triangle  for  common 
tangents. 

61.  Find  the  volume  and  also  the  inclination  of  two 
adjacent  faces  (1)  of  a  regular  tetraedron,  (2)  of  a  regular 
octaedron,  (3)  of  a  regular  dodecaedron,  and  (4)  of  a  regular 
icosaedron,  the  edge  being  one  inch. 

Ans.   (1)  117.85  cu.  in.,    70°31'43".4; 

(2)  .4714CU.  in.,  109°28'16"; 

(3)  7.663  cu.  in.,  116°  33' 54"; 

(4)  2.1817  cu.  in.,  138°  11'22".6. 

62.  In  the  tetraedron,  prove  (1)  that  the  circumradius 
is  equal  to  three  times  its  in-radius,  and  (2)  that  the  radius 
of  the  sphere  touching  its  six  edges  is  a  mean  proportional 
between  the  in-radius  and  circumradius. 

63.  Prove  that  th  ratio  of  the  in-radius  to  the  circum- 
radius is  the  same  in  the  cube  and  the  octaedron,  and  also 
in  the  dodecaedron  and  icosaedron. 


relations 


el  of  lati- 
l  over  by 
the  great 
ierence  of 

^sin  X)]. 

f  that  the 
itl  passing 
leet  at  it, 
r  common 

)n  of  two 
:  a  regular 
f  a  regular 

°31'43".4; 

°28'16"; 
i°33'54"; 
,°  11'  22".6. 

rcum  radius 
■j  the  radius 
roportional 


the  circum- 
n,  and  also 


